= Solution
\b[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.
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