A connected open subset of a normed space is polygonally connected.

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Theorem 7.4.2. A connected open subset of a normed space is path-connected.

Proof

Write $X$ for the connected open set. The key observation is that any path-component of $X$ is open. To see this, suppose that $P$ is a path-component of $X$, and let $a \in P$. Since $X$ is open, there is a ball $B(a, \varepsilon)$ contained in $X$. Let $b$ be a point in this ball. We can now write down an explicit path $\gamma$ between $a$ and $b$, namely $\gamma(t)=(1-t) a+t b$. This is easily seen to be continuous, and its image is contained in $B(a, \varepsilon)$ since
$$
\|\gamma(t)-a\|=t\|a-b\| \leqslant\|a-b\|=d(a, b)<\varepsilon
$$
for all $t$. Therefore $b$ lies in the same path-component $P$.
With this observation in place, the theorem follows easily. Indeed, the pathcomponents partition $X$, and so if there was more than one of them we could write $X$ as a disjoint union of non-empty open sets, contrary to the assumption that $X$ is connected.

The essential points in the proof are
• A ball in $X$ is path-connected.
so the result doesn't apply to $X=\{(x,y)\in\Bbb R^2:y=\sin1/x\}$ See this post (Theorem 7.4.3.)

• A path-connected component is an equivalence class.
so that we cannot replace "path-connected" with "convex"

We can strengthen the result by replacing "path-connected" with "polygonally-connected", the proof is similar:
A connected open subset of a normed space is polygonally-connected.
The following set is path-connected but can’t join 2 points by a polyline staying in the curve.