Does path-connected imply simple path-connected? Let $X$ be a path-connected topological space, i.e., for any two points $a,b\in X$ there is a continuous map $\gamma\colon[0,1]\to X$ such that $\gamma(0)=a$ and $\gamma(1)=b$. Note that beyond continuity little is required about $\gamma$.

Is it always possible to make $\gamma$ a simple (=injective) curve? (Of course we consider only the case $a\ne b$). 
  If the answer depends on $X$, what are some mild sufficient conditions on $X$?

If $X$ is sufficiently nice, e.g., an open subset of $\Bbb R^n$, the answer is "yes";  more generally, what one would call "locally simplepath-connected" suffices (the set of points reachable from $a\in X$ by a simple curve is both open and closed).
Also, for arbitrary $X$ it is clear that a single instance of self-crossing can be short-cut. But for infinitely many possible short-cuts the situation may become somewhat hairy, I'm afraid.
EDIT: To avoid Mike Miller's counterexample (indiscrete topology on $X$ with $|X|<|\Bbb R|$), let's at least assume that $X$ is Hausdorff. At any rate, $|X|\ge|\Bbb R|$ is a necessary condition - I am still looking for "as mild as possible"  sufficient conditions ...
 A: Since the OP asks for conditions, which are "mild as possible," I'll point out that the Hausdorff condition can actually be weakened slightly.
Definition: A space $X$ is $\Delta$-Hausdorff if the image of every path $\alpha:[0,1]\to X$ is closed in $X$.
We have the following.
Theorem: Every path-connected, $\Delta$-Hausdorff space is arcwise connected.
Proof. You can find the proof in this expository note.
At its heart, this proof is basically the one given in the accepted answer, where the idea is to make choices of loops to delete. Although the converse of the theorem is certainly false, I find it interesting that the following biconditional holds: a space $X$ is $\Delta$-Hausdorff if and only if the loop-deletion procedure can be applied to every non-loop path to produce an injective path.
It's also worth pointing out that it is possible to prove the theorem directly without using the axiom of choice. Such a proof is due to R. Borger, but this paper is probably difficult to come by if you don't have access to a University library.
The improvement is modest but properties like the $\Delta$-Hausdorff (or weakly Hausdorff) property are often used in algebraic topology and are strictly weaker than the Hausdorff property. For instance, the weakly Hausdorff property is a convenient separation axiom in the category $CG$ of compactly generated spaces. The $\Delta$-Hausdorff property is to the category of $\Delta$-generated spaces as the weakly Hausdorff property is to compactly generated spaces.
A: Following the hint given by mathcounterexamples.net, let me post a community-wiki translation (and rewording and elaboration) of a comment by JLT on a weblog by Pierre Bernard:
Theorem. Every path-connected Hausdorff space is simplepath-connected.
Before proving this theorem, we need some lemmas.
Lemma 1. Let $Y$ be a perfect (=no isolated points) compact metric space. Then there exists a diffuse probability measure on $Y$.
Proof.
It is known that such $Y$ contains a subspace homeomorphic to the Cantor set $C$. Moreover, the Lebesgue measure on $[0,1]$, transported to $C$ via the map $\sum a_k2^{-k}\mapsto \sum a_k3^{-k}$ (where $0.a_1a_2\ldots$ is the binary representation of $x\in[0,1)$) is a diffuse probability measure on $C$ and thus provides us with a diffuse probability measure on $Y$. $_\square$
Lemma 2. Let $Y$ be a perfect compact subset of a  metric space.
Then there exists a diffuse probability measure on $Y$ with support $Y$.
Proof.
For each $n$, $Y$ can be covered by a finite number $N(n)$ of open balls $B(x_{n,i},\frac1n)$, $1\le i\le N(n)$, of radius $\frac1n$. Let $Y_{n,i}=Y\cap\overline{B(x_{n,i},\frac1n)}$. The $Y_{n,i}$ are perfect compact metric spaces, hence by lemma 1 there exists a diffuse probability measure $\mu_{n,i}$ on $Y_{n,i}$. Then
$$\mu(A):=\sum_{n=1}^\infty\sum_{i=1}^{N(n)}\frac1{2^nN(n)}\mu_{n,i}(A\cap Y_{n,i}) $$
is a diffuse probability measure on $Y$. Its support intersects each $Y_{n,i}$ hence contains all of $Y$. On the other hand, the support is contained in the union of all $Y_{n,i}$, which is contained in $Y$. $_\square$
Recall that any open set $U\subseteq(0,1)$ can be written as disjoint union of countably many open intervals, $U=\bigsqcup_{i\in N}(a_i,b_i)$ with $N\subseteq\Bbb N$.
Definition. Let $f\colon [0,1]\to X$ be a curve. Let $U=\bigsqcup_{i\in N}(a_i,b_i)\subset [0,1]$ an open set. We say that $U$ is $f$-shortcutable if $f(a_i)=f(b_i)$ for all $i\in N$.
Lemma 3. Let $f\colon [0,1]\to X$ be a curve with $X$ Hausdorff. Let ${(a_i,b_i)}_{i\in I}$ be a chain (with respect to $\subseteq$) of $f$-shortcutable intervals. Then $\bigcup_{i\in I}U_i$ is $f$-shortcutable.
Proof.
We have $\bigcup_{i\in I}U_i=(a,b)$ with $a=\inf a_i$, $b=\sup b_i$.
Assume $f(a)\ne f(b)$ and let $V_a, V_b\subset X$ be open neighbourhoods separating $f(a)$ and $f(b)$. Then for some $\let\epsilon\varepsilon\epsilon>0$, $[a,a+\epsilon)\subseteq f^{-1}(V_a)$ and $(b-\epsilon,b]\subseteq f^{-1}(V_b)$.
Find $i,j\in I$ with $a_i\in [a,a+\epsilon)$, $b_j\in(b-\epsilon,b]$. Per chain property, $a_j\le a_i$ or $b_i\ge b_j$. In the first case $f(a_j)=f(b_j)\in V_a\cap V_b$, in the second case $f(b_i)=f(a_i)\in V_a\cap V_b$, contradiction. Hence $f(a)=f(b)$ and $(a,b)$ is $f$-shortcutable. $_\square$
Lemma 4.
Let $f\colon [0,1]\to X$ be a curve with $X$ Hausdorff.
Let $\{U_i\}_{i\in I}$ be a chain (with respect to $\subseteq$) of $f$-shortcutable open subsets of $(0,1)$. Then $\bigcup_{i\in I}U_i$ is $f$-shortcutable.
Proof. The connected components of the union are open intervals.
Consider one such component $(a,b)$ and a point $c\in(a,b)$. For $i\in I$ let $V_i$ be the component of $U_i$ containing $c$, or $V_i=\emptyset$ if $c\notin U_i$. Then the $V_i$ are a chain of $f$-shortcutable intervals.
By lemma 3, their union is an $f$-shortcutable interval $(a',b')\subseteq (a,b)$. If $a'> a$, then $a_i\in U_i$ for some $i$ and by openness  $[a',a'+\epsilon)\subseteq U_i$ for this $i$. As $a'+\frac\epsilon2\in V_j$ for some $j$, we conclude that $[a',a'+\epsilon)\cup V_j\subseteq U_k$ for some $k\in\{i,j\}$, which implies $a'\in(a',b')$, contradiction. We conclude $a'=a$. Similarly, $b'=b$. $_\square$
We are now finally ready to prove the theorem.
Proof of Theorem. Let $X$ be a Hausdorff space and $f\colon [0,1]\to X$ a path with $f(0)\ne f(1)$.
The set of $f$-shortcutable open subsets of $(0,1)$ is non-empty (contains $\emptyset$) and is inductively ordered according to lemma 4. By Zorn's lemma, there exists a maximal $f$-shortcutable $U_M$. As above write it as disjoint union of open intervals $$U_M=\bigsqcup_{i\in N}(a_i,b_i)$$ and then define
$$ x\sim y\iff x=y\lor\exists i\in N\colon \{x,y\}\subset [a_i,b_i].$$
This is an equivalence relation, where transitivity follows from maximality of $U_M$: Assume $x\sim y$ and $y\sim z$, say $\{x,y\}\in[a_i,b_i]$ and $\{y,z\}\in[a_j,b_j]$. If $i\ne j$ then from $(a_i,b_i)\cap (a_j,b_j)=\emptyset$ we conclude $y\in\{a_i,b_i\}\cap\{a_j,b_j\}$ (and specifically $y\notin U_M$); but then $f(a_i)=f(b_i)=f(a_j)=f(b_j)$ so that $U_M\cup\{y\}=U_M\cup(\min\{a_i,a_j\},\max\{b_i,b_j\})$ is $f$-shortcutable, contradicting maximality. Hence $i=j$ and clearly $x\sim z$.
The equivalence classes of $\sim$ are closed: In fact the class of $x$ is either $\{x\}$ alone, or a single interval $[a_i,b_i]$ or possibly the union of two adjacent closed intervals $[a_i,x]\cup [x,b_j]$ (though this possibility actually does not occur per maximality of $U_M$).
Let $J=[0,1]/{\sim}$. Being a quotient of the  compact Hausdorff space $[0,1]$ by an equivalence with closed equivalence classes, the space $J$ is also compact Hausdorff.
We can define a map $g\colon J\to X$  by letting $g([t]):=f(\inf [t])=f(\sup[t])$. This $g$  is injective and continuous. The theorem follows if we can show that $J\approx [0,1]$.
Let $Y_0=[0,1]\setminus U_M$. Then $Y_0$ is compact and has no isolated points except possibly $0$ and/or $1$. Let $Y$ be $Y_0$ minus its (at most two) isolated points (in other words, $Y=\overline{(0,1)\setminus U_M}$).
By lemma 2 there exists a diffuse probability measure $\mu$ on $[0,1]$ with support $Y$.
Consider $h\colon[0,1]\to [0,1]$, $x\mapsto \mu([0,x])$. This is a map  with $h(0)=0$ and $h(1)=1$; it is continuous because $\mu$ is diffuse. Moreover, for $x<y$
$$\begin{align}h(x)=h(y)&\iff \mu(\left]x,y\right[)=0 \\
&\iff\left]x,y\right[\cap\operatorname{supp}(\mu)=\emptyset\\
&\iff \left]x,y\right[\cap Y=\emptyset\\
&\iff \left]x,y\right[\cap Y_0=\emptyset\\
&\iff \left]x,y\right[\subseteq U_M\\
&\iff x\sim y \end{align}$$
Consequently, $h$ factors over $J$, thus showing $J\approx [0,1]$. $_\square$

Remark: The part where the homeomorphism $J\approx [0,1]$ is shown allows an alternative proof not involving the measure-theoretic lemmas 1 and 2 (which are not required for the proof of the theorem anywhere else):
We can define an order on $J$ by letting $[x]<[y]\iff x<y$, which is well-defined because the equivalence classes are in fact non-overlapping closed intervals. The topology of the quotient space $J$ is the order topology induced by this order because that is the case for $[0,1]$.
Also, $J$ is separable because $[0,1]$ is.
$J$ is complete: If $A\subset J$ is non-empty and bounded from above by $[x]\in J$, then it has a least upper bound; indeed, the preimage of $A$ in $[0,1]$ has a least upper bound $y\le x$ and then $[y]$ is a least upper bound for $A$. The same goes for lower bounds. The order is also dense as for $[x]<[y]$ we find $[x]<[z]<[y]$ with $z=\frac{\sup [x]+\inf[y]}2$, again because the equivalence classes are closed intervals.
In particular, $J$ does not consist only of the endpoints $[0]$ and $[1]$ so that $J\setminus\{[0],[1]\}$ is a nonempty, separable, complete, dense, endless total order and hence isomorphic to $\Bbb R$. Add back the two endpoints and we arrive at $[0,1]$ as desired.

Remark:
Invoking Zorn's lemma means invoking the Axiom of Choice.
As many prefer to avoid that axiom (if they have the choice), we can try to find a maximal shortcutable set more constructively:
We still assume $X$ Hausdorff.
Let $U$ be $f$-shortcutable and assume $U$ is not maximal.
Each $f$-shortcutable proper superset of $U$ has a component $(a,b)$ that is not a component of $U$. In other words,
the set
$$ A=\{\,(a,b):0\le a<b\le 1, f(a)=f(b), \{a,b\}\subsetneq [0,1]\setminus U\,\}$$
is non-empty. Let $n\in\Bbb N$ be minimal with
$$ A_n=\{\,(a,b)\in A:b-a\ge\tfrac 1n\,\}\ne\emptyset$$
and then let $\alpha =\inf\{\,a:(a,b)\in A_n\,\}$.
Claim. There exists $b$ with $(\alpha,b)\in A_n$.
Proof. Otherwise we find a strictly decreasing sequence $(a_k)$ with $a_k\to a$ together with a sequence $(b_k)$ such that $(a_k,b_k)\in A_n$. Then $(b_k)$ has an accumulation point $b\in [a+\frac1n,1]$.
Assume $f(a)\ne f(b)$ and as $X$ is Hausdorff let $V_a,V_b\subset X$ be open neighbourhoods separating $f(a)$ and $f(b)$. Then $f^{-1}(V_b)$ contains infinitely many $b_k$ and $f^{-1}(V_a)$ contains almost all $a_k$, hence for some $k$ we have $a_k\in f^{-1}(V_a)$, $b_k\in f^{-1}(V_b)$ and so $f(a_k)=f(b_k)\in V_a\cap V_b$, contradiction.
We conclude that
$f(b)=f(a)$. Clearly, $b-a\ge\frac 1n$. We have $a\notin U$, $b\notin U$ because all $a_k,b_k$ are $\notin U$. Finally, there are infinitely many $a_k\notin U$ between $a$ and $b$, and the claim follows. $_\square$
So we let $\beta=\sup\{\,b:(\alpha,b)\in A_n\,\}$. By continuity, $f(\beta)=f(\alpha)$ and ultimately we define $S(U)=U\cup (\alpha,\beta)\supsetneq U$. (For the sake of completeness, define $S(U)=U$ if $U$ is maximal).
For each ordinal $\nu$, we define an $f$-shortcutable set $U_\nu$ such that $\nu'<\nu$ implies $U_{\nu'}\subseteq U_\nu$. We do this  by the recursion $U_\nu=S(\bigcup_{\nu'<\nu}U_{\nu'})$. (Regarding the union, see lemma 4).
There must exist an ordinal $\nu$ with $U_\nu=U_{\nu+1}$, and this is a maximal $f$-shortcutable set as desired.
