I am given this problem:
Suppose that $\{C_{\alpha}\}$ is a family of connected subspaces of a space $X$. If each intersection $C_{\alpha} \cap C_{\alpha'}$ is nonempty, then $\cup_{\alpha} C_{\alpha}$ is connected.
And I am supposed to use the following lemma:
If $A \subseteq X$ is both open and closed, then any connected subspace $C \subseteq X$ that has a nonempty intersection with $A$ must be contained in $A$.
Could any of you offer me a hint as how to proceed?
Look at the contrapositive: suppose $X:=\cup_{\alpha}C_{\alpha}$ is disconnected, say $X=A\sqcup B$ where $A$ and $B$ are both nonempty and open. Note that both are both immediately closed as well. Then, each $C_{\alpha}$ must intersect either $A$ or $B$, and hence by your lemma be completely contained in $A$ or $B$. Also, there is at least one $C_{\alpha}$ intersecting $A$, and at least one $C_{\beta}$ intersecting $B$ (why?). Can you now show that we must have $C_{\alpha}\cap C_{\beta}=\varnothing$ (which is the result we wanted)?
If the union is not connected then it can be partitioned into two disjoint, non-empty, relatively open subsets $A$ and $B$. Let $x \in \cap C_{\alpha}$ and WLOG $x \in A$ and also $B$ is non empty. So take any element in $B$ , then there is some connected set in the union where the element belong. But that connected set has non empty intersection with both $A$ and $B$ which is a contradiction that the set is connected. Thus...