Is the complement of an injective continuous map $\mathbb{R}\to \mathbb{R}^2$ with closed image necessarily disconnected? I am interested in the following Jordan curve theorem-esque question:

Suppose that you are given a continuous, injective map $\gamma:
 \mathbb{R}\to \mathbb{R}^2$ such that the image is a closed subset of
  $\mathbb{R}^2$.  The complement of the image is an open subset of
  $\mathbb{R}^2$ and therefore consists of some number of path connected
  components.  Must this number always be greater than 1?

There are several cases in which this question reduces to the Jordan curve theorem, and one in which it does not (this last case is where the subtlety seems to lie):
Case 1: If the limits $\lim_{t\to \pm \infty} |\gamma(t)| = \infty$ then we can extend $\gamma$ continuously to a continuous, injective map $S^1\to S^2$ via stereographic coordinates. The result then follows by the Jordan curve theorem.
Case 2: If either of the limits $\lim_{t\to \pm \infty} \gamma(t)$ exist, then since the image of $\gamma$ is closed, the limit must equal $\gamma(t_0)$ for some $t_0 \in \mathbb{R}$. Then $\gamma$ induces a continuous, injective mapping $S^1 \to \mathbb{R}^2$, and the result follows by the Jordan curve theorem.
Case 3: If neither of the limits $\lim_{t\to \pm \infty} \gamma(t)$ exist, and $\lim_{t\to  \infty} |\gamma(t)| \neq \infty$ or $\lim_{t\to - \infty} |\gamma(t)| \neq \infty$, then it's unclear what may happen, but I can't imagine an example where the complement is connected. 
For example, consider a figure-8 type curve such that the ends approach the middle of the figure 8 from both sides like a topologists sine curve pictured below (this picture is edited from a picture in Munkres Topology, 2ed, p381, or the examples discussed in this answer: Is there a continuous injective map from $\mathbb{R}$ that has compact image?). The complement of the image of this curve has three components. 

 A: Here is a proof, which assumes familiarity with Chech cohomology and Alexander duality. For the latter, see A.Dold, "Lectures on Algebraic Topology", formula (8.18): If $X\subset {\mathbb R}^n$ is a closed subset, then 
$$
 \check{H}^{n-i}_c(X)\cong \tilde{H}_{i-1}({\mathbb R}^n -X).
$$ 
Suppose that $f: {\mathbb R}\to {\mathbb R}^n$ is an injective continuous map whose image is closed in ${\mathbb R}^n$. Then the restriction of the Euclidean metric from ${\mathbb R}^n$ to $Z:=f({\mathbb R})$ is complete; in particular, $Z$ (with the subspace topology) is a Baire space. 
Define the subsets $A_\pm\subset Z$ consisting of all points $a\in Z$ for which there exists a sequence $t_n\to  \pm\infty$ such that $\lim_{i\to\infty}f(t_i)=a$. Set $A= A_- \cup A_+$. Clearly, both $A_+, A_-$ are closed. 
The subset $A^c:= Z- A$ consists of all points $x=f(t)$ such that for some (equivalently, every) $r>0$, $x$ is an interior point of $f([t-r, t+r])$ in $Z$. 
Remark. The earlier version of my answer contained an error: I was assuming that $f([0,\infty))$ is closed which need not be the case. See also an edit at the end of this answer.   
Lemma 1. $A\ne Z$. 
Proof.  If not then for every $i\in {\mathbb N}$, $f([-i,i])$ has empty interior in $X$. Then $Z$ is a union of countably many subsets with empty interior. But $Z$ is a Baire space. A contradiction. qed
I will assume from now on that $A\ne \emptyset$, otherwise $f$ is a proper map and $f({\mathbb R})$ is homeomorphic to 
the real line. Without loss of generality (by precomposing $f$ with the mat $t\mapsto -t$), we can assume that $A_+\ne \emptyset$.  
Lemma 2. If  $A_+\ne \emptyset$ then there exists $x=f(t)\in A^c$ and $a=f(t_0)\in A_+$ such that $t > t_0$. 
Proof. If $f^{-1}(A_+)$ is unbounded from below, then any $x\in A^c$ will work. If (the closed subset) 
$f^{-1}(A_+)$ is bounded from below by some $t_0\in {\mathbb R}$ then the subset $f([t_0,\infty))$ is closed in ${\mathbb R}^n$. I will take $t_0$ to be the infimum of $f^{-1}(A_+)$. In particular, $f(t_0)\in A_+$. 
Now, the same argument as in the proof of Lemma 1 shows that $f([t_0,\infty))$ is not contained in $A_+$. Hence, 
there exists $t\in (t_0,\infty)$ such that $f(t_0)\notin A_+$. qed 
After precomposing $f$ with a translation of ${\mathbb R}$, we may assume that $t_0=0$. 
From now on, I will assume that $n=2$, that $A_+\ne \emptyset$, and, moreover, there exists a sequence $t_i\to \infty$ such that $\lim_{i\to\infty}f(t_i)=a=f(0)$. (See above.)  Fix a point  $b=f(t)\in A^c\subset X=f({\mathbb R}_+)$. 
Proposition. $\check{H}^1_c(X)\ne 0$. (Here and in what follows, I use only (co)homology with integer coefficients.) 
Proof. I will consider a sequence  of locally finite open covers ${\mathcal U}_n$ of $X$ by subsets $U_{k,n}$ such that
$$
\lim_{n\to\infty} \sup_{k} diam(U_{k,n})=0 
$$
and that for each $n$, ${\mathcal U}_{n+1}$ refines ${\mathcal U}_n$. 
Let $N_n$ denote the nerve of ${\mathcal U}_n$ with the vertex $v_{k}$ corresponding to the open set $U_{k,n}$. 
Since there exists an interval $I=(t+\delta, t-\delta)\subset {\mathbb R}_+$ such that $f(I)$ is open in $X$, without loss of generality we may assume that for each $n$, $f(t)$ belongs to the intersection $U_{1,n}\cap U_{2,n}$ of some 
 $U_{1,n}, U_{2,n}\in {\mathcal U}_n$ such that, moreover, 
 $$
 U_{1,n}\cap U_{2,n} \cap U_{k,n}=\emptyset
 $$
 for all $k\notin \{1, 2\}$. I then define a 1-cocycle $c\in Z^1({\mathcal U}_n)$ by 
$$
c(U_{1,n}\cap U_{2,n})=1,
$$
and $c(U_{j,n}\cap U_{k,n})=0$ otherwise. (Emptyness of the triple intersections above implies that this is indeed a cocycle.) In terms of the nerve $N_n$, the cocycle $c$ is defined by $c([v_1, v_2])=1$ and $0$ for all other edges. 
I claim that  this cocycle is nontrivial (when $n$ is sufficiently large). I will prove its nontriviality by exhibiting a 1-cycle $\sigma\in Z_1(N_n)$ such that $\langle c, \sigma\rangle =1$. Namely, for each $n$ there exists $y_i=t_i$ such that $f(t_i), a$ belong to a common $U_{k,n}\in {\mathcal U}_n$. By concatenating $f([0, t_i])$ with the line segment $y_i a$ we obtain a (likely nonsimple)  loop  $\lambda_n$ in ${\mathbb R}^2$. This loop defines a 1-cycle $\sigma$ in $N_n$ as follows. Pick points $s_0=0\le s_1\le s_2\le ... \le s_{p}=t_i$ in $[0, t_i]$ such that for every $j$, $\{f(s_j), f(s_{j+1})\}\subset U_{q_j,n}\in {\mathcal U}_n$. We set $U_{q_p,n}=U_{q_0,n}$. We obtain a simplicial 
loop $\sigma$ in $N_n$ with the vertices 
$$
v_{q_0}, v_{q_1},..., v_{q_{p-1}}
$$
Moreover, assuming that 
$$
\max_{j} |s_{j+1}- s_j| 
$$
is sufficiently small, we achieve that the edge $[v_{1,n}, v_{2,n}]$ appear exactly one in this loop $\sigma$. Hence
  $\langle c, \sigma\rangle =1$. 
By the construction, the natural map $\kappa_n: N_{n+1}\to N_n$ satisfies
$$
\kappa_n^*(c_n)= c_{n+1}. 
$$
Thus, the sequence of 1-cycles $(c_n)$ defines a nonzero element of 
$$
\lim_{n} H_c^1({\mathcal U}_n)= \check{H}^1_c(X).  
$$
qed
We can now finish the proof:
Lemma 3. $X=f([0,\infty))$ separates ${\mathbb R}^2$. 
Proof. By the Alexander duality,
$$
 \check{H}^1_c(X)\cong \tilde{H}_0({\mathbb R}^2 -X).
$$
Since $\check{H}^1_c(X)\ne 0$, so is $\tilde{H}_0({\mathbb R}^2 -X)$. qed 
We now are ready to prove:
Theorem. Suppose that $f: {\mathbb R}\to {\mathbb R}^2$ is a continuous injective map with closed image. Then $Z:=f({\mathbb R})$ separates ${\mathbb R}^2$. 
Proof. Define the subset $A=A_+ \cup A_-\subset Z$ as before. Then $A=\emptyset$ iff $f$ is a proper map. In this case, by the Alexander duality,
$$
{\mathbb Z}\cong \check{H}^1_c(Z)\cong \tilde{H}_0({\mathbb R}^2 -Z),
$$
hence, $Z$ separates ${\mathbb R}^2$. If $A\ne \emptyset$, then, as noted earlier, without loss of generality, we may assume that $A_+\ne \emptyset$. Then Lemma 3 implies that ${\mathbb R}^2- (X=f([0,\infty)))$ is a disjoint union $U\sqcup V$ of two nonempty open subsets. By applying Baire category argument as in Lemma 1, we see that neither $U$ nor $V$ is contained in $Z$. Hence, ${\mathbb R}^2 - Z= (U - Z)\sqcup (V- Z)$ is a disjoint union of two nonempty open subsets and, hence, is not connected. qed 
Edit. Although we do not need this, here is an interesting fact about the set $A_+$. Define the subset $J:= {\mathbb R} - f^{-1}(A_+)$. It is easy to construct examples where $J$ is bounded below (using a version of the topologist's sine curve). However, we have:
Lemma 4. 
$J\subset {\mathbb R}$ is unbounded above. 
Proof. The subset $A^c_+:=f(J)\subset Z$ is open, hence, its complement $Z-  A^c_+= A_+$ is closed. Assume for a moment that $J$ is bounded above. Then removing $J$ from ${\mathbb R}$ does not affect the accumulation set $A_+$. It follows that each $a\in Z  - A^c_+$ equals the limit 
$$
a=\lim_{t_i\to \infty} f(t_i). 
$$
Therefore, for every interval $[-n,n]$, the subset $f([-n,n] -J)$ has empty interior in $A_+$. Thus, we obtain a contradiction with the Baire property as in the proof of Lemma 1. qed 
A: Here is an incomplete solution that answers the question in the affirmative in a subcase of case 3.


*

*Suppose that $\lim_{t\to \infty} \gamma(t)$ DNE.

*Let $S = \{ x\in \mathbb{R}^2 : \exists y_i \to \infty \text{ such that } \gamma(y_i) \to x\}$, and assume it is non-empty (e.g. if $|\gamma(t)|$ is bounded as $t\to \infty$).

*The set $S$ is closed and contained in the image of $\gamma$ (by assumption that the image is closed), so the preimage $\gamma^{-1}(S)$ is closed and nonempty.


It is not clear whether the set $\gamma^{-1}(S)$ is necessarily bounded. However, if we assume that this set is bounded, we can prove that the complement of the image of the curve has at least two connected components (and the argument is similar to the argument that the complement of the closed topologists sine curve is disconnected):


*

*If $\gamma^{-1}(S)$ is closed and bounded, then it has a maximum element $x$.

*Let $y$ be a number larger than $x$ and consider the following two closed sets: $A_1$ is the image under $\gamma$ of the segment $[x,y]$, and $A_2 = \gamma([y,\infty)) \cup S$ is the closure of the image of the set $[y,\infty)$. By definition, $A_i \subset$ the image of $\gamma$.

*The two sets $A_i$ are closed by definition and their intersection consists of exactly two points, $\gamma(x)$ and $\gamma(y)$, thus by "a general separation theorem" (Munkres, 2ed, page 380, Theorem 61.4), the complement of $A_1 \cup A_2$ is disconnected. Thus the complement of the image of $\gamma$ is disconnected.


Now, this does not appear to be a complete proof that the answer to the question above is yes: it seems conceivably possible that the set $\gamma^{-1}(S)$ above is not bounded (imagine that in the limit $t\to \infty$ the curve $\gamma$ follows itself, oscillating faster and faster). Here is an illustration:

It might also be possible that $S$ is empty to begin with (although I can't imagine an example of a curve with this property).
A: I think I have a (tricky, fiddly) proof of this under a particular assumption - to make this assumption I'll need the following definition:
Definition Let $A \subset \mathbb{R}$ be the set of points $a$ such that there is a sequence $a_1, a_2, ...$ in $\mathbb{R}$ with $\lim_{i\rightarrow \infty}a_i = \infty$ and $\lim_{i\rightarrow \infty}\gamma(a_i) = \gamma(a)$.
The assumption I'll need is the following:
Assumption $A$ does not contain any interval of the form $(x, \infty)$, ie the complement of $A$ has no upper bound.
Edit:  I just saw Moishe Cohen's answer, which starts by defining $A$ in a similar way, and proves that $A$ is has nonempty complement using the fact that a closed subset of $\mathbb{R}^n$ is a Baire space. I believe this argument also shows that $A$ has no upper bound, proving my assumption.
Lemma 1 In case 3, $A$ is non-empty.
Proof Since $\lim_{x\rightarrow\infty} \neq \infty$, there is some bounded region $R \subset \mathbb{R}^2$ for which $\gamma(x) \in R$ for arbitrarily large $x$.  Take an unbounded sequence of such $x$, look at their images under $\gamma$ and find a convergent subsequence (which must exist because $R$ is compact).  The limit of this sequence must lie on the curve since the image of $\gamma$ is closed, and is thus in $A$. $\square$
In light of this lemma and the assumption, we can choose points $a,b \in \mathbb{R}$ such that $b > a$, $a \in A$, and $b \notin A$.
Lemma 2 For any $x > b$, there exists an $\epsilon > 0$ such that for any $x' \geq x$, $d(\gamma(x'),\gamma(b)) > \epsilon$ (where $d$ is just the standard distance metric on $\mathbb{R}^2$).
Proof Suppose otherwise.  Then we can find an unbounded sequence of $x_i$ in $\mathbb{R}$ whose images approach $\gamma(b)$, contradicting the fact that $b \notin A$. $\square$
Next, a bunch of fiddly construction:
Let $s$ be the distance between $\gamma(a)$ and $\gamma(b)$. Since $a \in A$, we can choose a $y_0 > b$ such that $d(\gamma(a), \gamma(y_0)) < \frac{s}{3}$.  Choose $z_0 \in [b, y_0]$ that minimizes $d(\gamma(z_0), \gamma(a))$.  Draw the line segment $s_0$ from $\gamma(z_0)$ to $\gamma(a)$, and let $a_0$ be the intersection of $\gamma$ and $s_0$ closest to $\gamma(z_0)$ (not counting $\gamma(z_0)$ itself.
Note: The length of $s_0$ is at most $\frac{s}{3}$, and every point on $s_0$ is at most $\frac{s}{3}$ from $a$.
Let $\gamma_0$ be a closed loop in $\mathbb{R}^2$ defined to first follow $\gamma([a_0, z_0])$ then to follow $s_0$ from $\gamma(z_0)$ back to $\gamma(a_0)$.
Note: We have carefully constructed $\gamma_0$ to be a genuine loop with no self-intersections, thus it splits the plane into two path components of which $\gamma_0$ is the boundary of both.
Using Lemma 2, choose an $\epsilon > 0$ such that for any $x' > y_0$, $d(\gamma(x'), \gamma(b)) > \epsilon$. Pick two points $p_1$ and $p_2$, one in each path component, each within $\epsilon$ of $\gamma(b)$. Draw a path from each of them to $\gamma(b)$ that is otherwise disjoint from $\gamma_0$, and by replacing $p_1$ and $p_2$ with points far enough along these paths we can assume they are both connected to $\gamma(b)$ by a path disjoint from $\gamma$ that is entirely within $\epsilon$ of $b$, and the two paths approach $\gamma(b)$ from opposite sides (ie different local path components of a neighborhood of $\gamma(b)$ in $\mathbb{R}^2 - \gamma$.
Now, we assume for contradiction that $\mathbb{R}^2 - \gamma$ is connected.
This means we can draw a path $\beta$ connecting $p_1$ and $p_2$ in $\mathbb{R}^2 - \gamma$.  Since we know $p_1$ and $p_2$ are in separate components of $\mathbb{R}^2 - \gamma_0$, this path must cross $\gamma_0$, and in particular must intersect $s_0$, which means it passes within $\frac{s}{3}$ of $\gamma(a)$.
Now take an unbounded sequence of values $y_0 < y_1 < y_2 < y_3 < ...$ such that $\lim_{i\rightarrow\infty}\gamma(y_i) = \gamma(a)$ and such that $d(\gamma(y_{i+1}),\gamma(a)) < d(\gamma(y_i), \gamma(a))$.  Repeat the process used to construct $z_0, a_0, s_0,$ and $\gamma_0$ from $y_0$ to construct $z_i, a_i, s_i$ and $\gamma_{i}$ from $y_i$.
Note: that $p_1$ and $p_2$ lie in separate path components of $\mathbb{R}^2 - \gamma_i$ for each $i$, since the $\gamma_i$ all coincide on a neighborhood of $\gamma(b)$ containing $p_1$ and $p_2$ and the paths connecting them to $\gamma(b)$.
Note: The lengths of the segments $s_i$ approach zero, as do the distances from $s_i$ to $a$.
This means the path $\beta$ must intersect each segment $s_i$, but the entirety of the $s_i$ get arbitrarily close to $\gamma(a)$.  Thus, the sequence of intersections of $\beta$ with the $s_i$ converges to $\gamma(a)$.  Since $\beta$ is a closed path this implies $\gamma(a)$ lies in the image of $\beta$, contradicting the fact that $\beta$ was a path in $\mathbb{R}^2 - \gamma$. $\square$
Finally, I'll note that the assumption I make for this proof doesn't hold for all paths - AnonymousCoward provides such an assumption-breaking path in their answer.  However, I'm pretty convinced that all paths which break this assumption don't have closed image.
