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.