Proposition 4.12.2. Let $X$ be a topological space, $\seqi{A}\subset 2^{X}$ be connected with $\bigcap_{i \in I}A_{i} \ne \emptyset$, then $\bigcup_{i \in I}A_{i}$ is connected.
Proof. Let $x \in \bigcap_{i \in I}A_{i} = A$ and $U, V \subset X$ be open with $U \cap A, V \cap A \ne \emptyset$. Assume without loss of generality that $x \in U$. Let $i \in I$ such that $V \cap A_{i} \ne \emptyset$, then since $U \cap A_{i} \ne \emptyset$, $U \cap V \ne \emptyset$ by connectedness of $A_{i}$.$\square$