 Solution

True. Fix a point $x_0$ of a nonempty connected open set and let $P$ be its path-connected component. Each point of the open set has a small Euclidean ball contained in it. A ball is convex and path-connected, so each such ball lies entirely in one path component. All path components are therefore open, and the complement of $P$ is open as well. If the complement were nonempty, these two open sets would disconnect the original set. Thus $P$ is the whole set. In fact the joining paths may be chosen polygonal.