Open connected subsets of R^n are path-connected

Statement

Let $n \ge 1$ and let $U \subseteq \mathbb{R}^n$ be open and connected, for the Euclidean topology. Then $U$ is path-connected or empty.

Remarks

Proof: fix $x \in U$, and let $P$ be the set of points of $U$ that a path in $U$ joins to $x$. Each open ball in $U$ is convex, so $P$ and $U \setminus P$ are both open. As $U$ is connected and $x \in P$, we get $P = U$. The proof works in each locally path-connected space, for example in each open subset of a normed space. In fact, any two points of $U$ can be joined by a path made of finitely many line segments.

Explanations

L1 human

For open sets in $\mathbb{R}^n$, being in one piece and being walkable are the same. The bad behaviour of the topologist's sine curve cannot happen in an open set, because each point has a small ball around it.