Can we divide $\mathbb{R}^2$ into two path connected parts such that each part is not simply-connected? 
Can we divide $\mathbb{R}^2$ into two connected parts such that each part is not simply-connected?

My attempt
Put $A= \{ (0,0) \} $ and $B$ is the punctured plane.
Since that $S^1$ is a deformation retract of the punctured plane, $B$ is not simply-connected. Thus we can find a division of $\mathbb{R}^2$ such that one part is simply-connected but the other is not.
But how to deal with the problem above which requires that each part is not simply-connected? 
It seems to be related to contractible and holes. But I don't know how to convert these ideas into precise mathematical language.
Any hints? Thanks in advance!
Added:
As pointed out in the comment, the counterexample exists.
Now I want to ask another question

Can we divide $\mathbb{R}^2$ into two path connected parts such that each part is not simply connected?

 A: Both $\{ (x,\sin (1/x)) \mid x \neq 0 \} \cup \{(0,0)\}$
and it's complement are connected.
Since they are not path connected,
they are not simply connected.  
It is an example of two connected sets that pass through each other.  
Are there two path connected, not simply connected sets whose union is the plane?
A: The answer to the modified question:

Can we divide $\mathbb R^2$ into two path connected parts such that each part is not simply connected?

is no, this is not possible. In fact, if $A$ and $B$ are disjoint sets with $A\cup B=\mathbb R^2$ then, if $A,B$ are path connected but $A$ is not simply connected, then $B$ must be bounded. So, if $A$ and $B$ are both path connected but not simply connected, then they must both be bounded, giving a contradiction.
I will make use of the following lemma, from the paper The fundamental groups of subsets of closed surfaces inject into their first shape groups.

Lemma 13: Let any set $A\subseteq\mathbb R^2$, and map $\alpha\colon S^1\to A$ be given. Let $U$ be the unbounded connected component of $\mathbb R^2\setminus{\rm Im}(\alpha)$. If $\alpha\colon S^1\to A$ is null-homotopic, then so is $\alpha\colon S^1\to A\setminus U$.

So, suppose $A$ is connected but not simply connected. Then there is a curve $\alpha\colon S^1\to A$ which is not null-homotopic. The complement $\mathbb R^2\setminus {\rm Im}(\alpha)=\bigcup_iU_i$ decomposes as the union of its connected components $U_i$. As $B\subseteq\bigcup_iU_i$ is connected, it will be contained within one of the $U_i$. We show that $U_i$ is bounded: As $\mathbb R^2\setminus U_i\subseteq A$, $\alpha$ is not null-homotopic in $\mathbb R^2\setminus U_i$. If $U_i$ were unbounded, then the quoted result says that $\alpha$ is not null-homotopic in $\mathbb R^2$, a contradiction.
A: So here's my partial answer. Assume additionally that

In each space there is a Jordan curve that is not null homotopic.

Let's assume $\lambda$ is such curve in $X$. With that we apply the Jordan curve theorem. Say $\mathcal{I}$ is the interior of $\lambda$ and $\mathcal{E}$ the exterior. If $\mathcal{I}\cap Y\neq\emptyset$ and $\mathcal{E}\cap Y\neq\emptyset$ then $Y$ cannot be path connected because any path (between exterior and interior) would have to cross $\lambda$.
On the other hand if $\mathcal{I}\cap Y=\emptyset$ then $\mathcal{I}\subseteq X$ and so $\lambda$ is null homotopic in $X$ because by the Jordan–Schoenflies theorem the interior of $\lambda$ is homeomorphic to a ball.
So the only possibility is that $\mathcal{E}\cap Y=\emptyset$, i.e. $\mathcal{E}\subseteq X$.
The same reasoning can be applied to $Y$ in order to get that the exterior of some Jordan curve has to be fully contained in $Y$. This contradicts them being disjoint.
So I couldn't find an example of a path-connected, not simply-connected subset of $\mathbb{R}^2$ which doesn't have a not null homotopic Jordan curve. But I also couldn't prove that this is not possible. Any help appreciated. Note that such subsets do exist in $\mathbb{R}^3$ (see here).
Side note: If we don't require that both have to be path-connected, but at least one, then there's a neat counterexample of $X=\mathbb{Q}^2$ and $Y=\mathbb{R}^2\backslash\mathbb{Q}^2$. Note that $Y$ is path-connected but not simply connected.
A: It will be very difficult to find an example (and I do not believe it is possible). Here is a partial result.
Let $X \subset \mathbb R ^2$ be compact and path connected. If its complement $\mathbb R ^2 \setminus X$ is connected (which means that $X$ does not separate the plane) , then $X$ is simply connected.
This relies on some deep theorems, and I can only give an outline. The paper
Hagopian, Charles. "The fixed-point property for simply connected plane continua." Transactions of the American Mathematical Society 348.11 (1996): 4525-4548.
contains the following results:
Theorem 1.2. Every arcwise-connected nonseparating plane continuum has the fixed-point property.
Theorem 9.4. Suppose $M$ is an arcwise-connected plane continuum. For $M$ to
have the fixed-point property it is necessary and sufficient that the fundamental
group of $M$ be trivial.
The above claim is an immediate consequence.
I could imagine that there is a more direct proof (which does not focus on the fixed point property), but I don't have any idea how to do that.
Remark: One can moreover show that if $X$ is a compact connected subset of the plane such that $\mathbb R ^2 \setminus X$ is connected (which is the same as path connected because it is an open set), then $\mathbb R ^2 \setminus X$is homeomorphic to $\mathbb R ^2 \setminus \{ 0 \}$. See my answer to  Homotopy type of the complement of a subspace.
A: If there is an example then it is nasty. With some mild hypotheses it's not possible. For notation consider $S^2 = \mathbb{R}^2 \cup \{\infty\}$.
Suppose that $\mathbb{R}^2 = X \cup Y$ and that $X$ is path connected but not simply connected. Suppose in addition that $X$ is a compact CW-complex. Then it deformation retracts onto a graph, so its fundamental group is free (and nontrivial by hypothesis), thus its first cohomology group is nontrivial. Then by Alexander duality, $\tilde{H}_0(S^2 \setminus X) \cong \tilde{H}^1(X) \neq 0$.
Note that $S^2 \setminus X = Y \cup \{\infty\}$. Since $X$ is compact, it follows that $Y$ is a neighborhood of $\infty$. Removing a point from an open subset of $S^2$ does not change its connectivity. It follows that $\tilde{H}_0(Y) = \tilde{H}_0(S^2 \setminus X) \neq 0$ and therefore $Y$ is not path-connected.
There is probably a way around the hypothesis that $X$ is compact (if $X$ is noncompact then add $\infty$ to it and then $S^2 \setminus X = Y$). However I'm not sure how to get rid of the assumption that $X$ is nice; it has to be at least locally contractible to apply Alexander duality. Perhaps one can work something out with Čech cohomology, but then I don't know how to see that $\check{H}^1(X) \neq 0$.
A: I believe the answer to the question 
"Can we divide $\mathbb{R}^2$ into two path connected parts such that each part is not simply connected?" is false.
Let the two (disjoint) parts be A and B. I claim that if A is not simply connected then B is not path connected:
If A is not simply connected then there exists a simple closed loop L lying entirely in A s.t L cannot be contracted to a point. This means that the region R surrounded by L must contain a "hole", more precisely there exists a point $p\in R$ which does not lie in A and hence must lie in B. This means that B cannot be path connected since any path that starts at p must cross L (by the Jordan Curve Theorem) and hence can not lie entirely in B.
UPDATE: The answer to this question "In a subset of $\mathbb{R}^2$ which is not simply connected does there exist a simple loop that does not contract to a point?" that I posed on MathOverflow indicates that it is probably true that in a subset of $\mathbb{R}^2$ that is not simply connected there must exist a simple loop that contracts to a point. Therefore this makes the above proof probably true. Note that this is a special property of $\mathbb{R}^2$ since, as freakish points out, there exist spaces in $\mathbb{R}^3$ that are not simply connected but where all simple closed curves contract to a point.
