TITLE: A kind of converse to the Jordan curve theorem
QUESTION [11 upvotes]: The Jordan curve theorem in $\mathbb{R}^2$ says that if $S$ is a closed curve in $\mathbb{R}^2$. Then $S$ splits $\mathbb{R}^2$ into exactly two connected components $A$ and $B$.
I was thinking about a kind of converse to this problem which is as follows.
Let $S$ be a closed and bounded set in $\mathbb{R}^2$ and let $x,y\in\mathbb{R}^2 \setminus S$. Define 
$$A:=\{a\in\mathbb{R}^2 \setminus S: a\text{ and } x \text{ are path connected in } \mathbb{R}^2\setminus S\}$$
$$B:=\{b\in\mathbb{R}^2\setminus S: b\text{ and } y\text{ are path connected in }\mathbb{R}^2 \setminus S\}$$
If $A\cap B=\emptyset$ does there exists a connected $T\subset S$  such that $\mathbb{R}^2 \setminus T$ is split into exactly two connected components one which contains $A$ and the other that contains $B$.
Any help or references on this would be much appreciated
Note: Ive made an edit to reflect the comments using the topologists sine curve. I believe that now this will not be a counter-example because the 'quasi-circle" is obviously connected. Also I apologize for not being more specific originally.

REPLY [10 votes]: Consider this variant of the extended topologist's sine curve:

(picture borrowed from this question)
It is the closure of the graph of the function $x\mapsto \sin(\pi/x)$ defined on the half open interval $(0,1]$ with the points $(0,0)$ and $(1,0)$ connected by a simple arc. It is a closed and bounded subset $S$ of $\mathbf R^2$. Its complement has exactly $2$ path-wise connected components, but there is no simple closed curve $T$ contained in $S$.

REPLY [7 votes]: Here's a counterexample which I think works. 
$S \subset \Bbb R^2$ be the subspace consisting of points $(x, \sin(1/x))$ for $x \in (0, 1]$ and points in $\{0\} \times [-1, 1]$ union an arc joining $(0, 0)$ with $(1, \sin(1))$, namely, the quasi-circle. 
$S$ is a closed and bounded subset of $\Bbb R^2$. One can see that $S$ separates $\Bbb R^2$ into two disjoint components by noting that image of any path $\gamma$ not hitting $S$ has a positive distance from $S$, as both are compact subspaces of $\Bbb R^2$. Then one can - with care - replace the $\{0\} \times [-1, 1]$ part of $S$ (because $\gamma$ has distance at least $\delta > 0$ from it) to turn it into a circle and invoke Jordan curve theorem.
However, there is no such $T$: any closed loop in $S$ must miss the origin $(0, 0)$ as $S$ is not path connected at that point. But then obviously such loops do not separate $\Bbb R^2$.