I believe that the following statement is true:
Let $E$ be a connected open subset of $\mathbb{R}^2$. For any $n$ distinct points in $E$, there exists a connected and simply connected open set $G \subset E$ that contains those points.
How can I show this? My original idea was to try to show this by induction, by arguing that there is a simple curve from the $n-1$-th point to the $n$-th point that doesn't intersect the set $G_{n-1}$ more than once.
But then, I realized I don't know how to show that given 2 points in an open connected set, there is a simple curve connecting the two. I thought that given any curve connecting two points, one could construct a simple curve, but I wasn't sure of how to deal with the case when there are infinitely many self-intersections.
Note: The statement I'm trying to show is equivalent to showing that given $n$ distinct points, there is a simple curve that connects the $n$ distinct points. But unfortunately I don't know how to show this either.