Does every simply-connected open subset of $S^2$ have connected complement?

Question

Given a simply-connected open subset $U \subset S^2$, is $S^2\setminus U$ connected?

Here is what I understand so far:

• The answer becomes "no" if the "open" condition is dropped, or replaced by "closed". E.g., take $U$ to be a Warsaw circle in $S^2$.
• The answer becomes "no" if the question ("is $X = S^2\setminus U$ connected") is replaced by ("is $X$ path-connected"). E.g., take $X$ to be a (closed) topologist's sine curve.

However I'm not sure how else to analyze this question. Would anyone have any suggestions?

Answer 1

Someone told me about a proof outline which proceeds along the following lines:

1. Connected components are always closed subsets;

2. Removing two (or more) nonempty disjoint closed subsets of $S^2$ produces a space which is not simply-connected.

Since path components need not be closed, this fails if "connected" had been replaced by "path connected".