TITLE: Union of connected subsets is connected if intersection is nonempty
QUESTION [35 upvotes]: Let $\mathscr{F}$ be a collection of connected subsets of a metric space $M$ such that $\bigcap\mathscr{F}\ne\emptyset$. Prove that $\bigcup\mathscr{F}$ is connected.

If $\bigcup\mathscr{F}$ is not connected, then it can be partitioned into two disjoint, non-empty subsets $A,B$. Let $x$ be a point in $\bigcap\mathscr{F}$. Then either $x\in A$ or $x\in B$. I don't know where to go from here.

REPLY [7 votes]: I like "clopen" sets... it is a set that is at the same time closed and open. A topological space $X$ is not connected if, and only if, for any point $a \in X$, you have a "non-trivial" clopen set $C$ containing $a$. That is, a clopen set such that $\emptyset \neq C \subsetneq X$.
Take $a \in \bigcap \mathscr{F}$. Then, take a clopen set $C \subset \bigcup \mathscr{F}$ containing $a$. In the relative topology, for any $X \in \mathscr{F}$, $C \cap X$ is a clopen set. Since $X$ is connected, $C \cap X = X$. That is, $X \subset C$ for every $X \in \mathscr{F}$. Therefore, $\bigcup \mathscr{F} \subset C$.
Therefore, $\bigcup \mathscr{F}$ has no "non-trivial" clopen set, and so, it is connected.

The proof admits some variants. For example, if you take any non-empty clopen set $C'$, it will intersect some $X \in \mathscr{F}$. From connectedness, we have that $X \subset C'$. In particular, $a \in C'$.