Proof. Let $U, V \subset [0, 1]$ be open with $[0, 1] = U \cup V$. If $\sup U = \sup V$, then $\sup U = \sup V = 1$ and $U, V \in \cn^{o}(1)$, so $U \cap V \ne \emptyset$. If $\sup U < \sup V \le 1$, then $x \not\in U$ and $x \in V$. In which case, $V \in \cn^{o}(x)$ and $V \cap U \ne \emptyset$. Therefore $[0, 1]$ is connected.

Fix $x \in X$. For any $y \in X$, let $f_{y} \in C([0, 1]; X)$ be a path from $x$ to $y$, then $f_{y}([0, 1])$ is connected with $x, y \in f_{y}([0, 1])$ by Proposition 4.12.3. By Proposition 4.12.2,

is connected.$\square$