Proof that open connected subsets of R^n are path-connected, by polygonal paths

Proof

We show that the set of points reached from a fixed point by polygonal chains is open, closed in $U$ and nonempty.

1. Write $\|v\|$ for the Euclidean norm on $\mathbb{R}^n$, the 2-norm, so that the Euclidean metric is $\|x - y\|$. For $p, q \in \mathbb{R}^n$, the segment $[p, q]$ is the set of points $(1 - t)p + tq$ with $0 \le t \le 1$.
2. If $U$ is empty, there is nothing to prove. Let $U$ be nonempty and fix $x \in U$.
3. A chain in $U$ from $x$ to $y$ is a finite sequence $p_0 = x, p_1, \ldots, p_m = y$ of points, with $m \ge 0$ and $[p_{k-1}, p_k] \subseteq U$ for $1 \le k \le m$. Let $P$ be the set of $y \in U$ with such a chain.
4. Each open ball $B(p, r)$ is convex. For $q \in B(p, r)$ and $0 \le t \le 1$, we have $\|(1 - t)p + tq - p\| = t \|q - p\| < r$.
5. $x \in P$, by the chain with $m = 0$.
6. $P$ is open. Let $p \in P$. As $U$ is open, the ball criterion gives $r > 0$ with $B(p, r) \subseteq U$. For $q \in B(p, r)$, step 4 gives $[p, q] \subseteq U$. So appending $q$ to a chain for $p$ gives a chain for $q$, and $B(p, r) \subseteq P$.
7. $U \setminus P$ is open. Let $p \in U \setminus P$ and $B(p, r) \subseteq U$. If some $q \in B(p, r)$ were in $P$, then $[q, p] \subseteq U$ by step 4. Appending $p$ to a chain for $q$ would put $p$ in $P$.
8. So $B(p, r) \subseteq U \setminus P$, and $U \setminus P$ is open.
9. The sets $P$ and $U \setminus P$ are open in $\mathbb{R}^n$ and lie in $U$, so they are open in the subspace topology of $U$. They are disjoint, and their union is $U$.
10. $U$ is connected and $P \neq \varnothing$. By the definition of a connected space, $U \setminus P = \varnothing$. So $P = U$.
11. Let $y \in U$ and take a chain $p_0, \ldots, p_m$ from $x$ to $y$. If $m = 0$, then $y = x$ and the constant map is a path. Let $m \ge 1$.
12. Define $\gamma \colon [0, 1] \to \mathbb{R}^n$ as follows. For $1 \le k \le m$ and $t \in [\tfrac{k-1}{m}, \tfrac{k}{m}]$, let $\gamma(t) = p_{k-1} + (mt - k + 1)(p_k - p_{k-1})$.
13. At $t = k/m$, the formulas for the pieces $k$ and $k + 1$ both give $p_k$, so $\gamma$ is well defined. Also $\gamma(0) = x$, $\gamma(1) = y$, and $\gamma(t) \in [p_{k-1}, p_k] \subseteq U$.
14. Let $L = m \max_k \|p_k - p_{k-1}\|$. If $s, t$ lie in the same piece $k$, then $\|\gamma(t) - \gamma(s)\| = m |t - s| \, \|p_k - p_{k-1}\| \le L |t - s|$.
15. If $s < t$ lie in different pieces, let $s < b_1 < \cdots < b_j < t$ be the points $k/m$ between them. The triangle inequality and step 14 on each subinterval give $\|\gamma(t) - \gamma(s)\| \le L (t - s)$.
16. So $\gamma$ is Lipschitz with constant $L$. Given $\varepsilon > 0$, the choice $\delta = \varepsilon / (L + 1)$ works in the epsilon-delta condition. By continuity via open sets, $\gamma$ is continuous into $\mathbb{R}^n$.
17. The open sets of $U$ are the sets $U \cap O$ with $O$ open in $\mathbb{R}^n$. Since $\gamma$ takes values in $U$, $\gamma^{-1}(U \cap O) = \gamma^{-1}(O)$ is open. So $\gamma$ is a path in $U$ from $x$ to $y$.
18. The point $x \in U$ was arbitrary. So any two points of the nonempty set $U$ are joined by a path in $U$, and $U$ is path-connected.