Proposition 4.13.4. Let $X$ be a topological space, $\seqi{A}$ be path-connected sets with $\bigcap_{i \in I}A_{i} \ne \emptyset$, then $\bigcup_{i \in I}A_{i}$ is connected.
Proof. Let $x \in \bigcap_{i \in I}A_{i}$, $i, j \in I$, $y \in A_{i}$, and $z \in A_{j}$. By connectedness of $A_{i}$ and $A_{j}$, there exists paths $f \in C([0, 1]; X)$ from $y$ to $x$, and $g \in C([0, 1]; X)$ from $x$ to $z$. Thus the concatenation
\[g \cdot f: [0, 1] \to X \quad t \mapsto \begin{cases}f(2t)&t \in [0, 1/2] \\ g(2(t - 1/2))&t \in [1/2, 1]\end{cases}\]
is a path from $y$ to $z$.$\square$