$\textbf{Question:}$ If $\{E_{\alpha} \}_{\alpha \in A}$ is a collection of connected subsets of $X$ such that $\bigcap_{\alpha \in A} E_{\alpha} \neq \emptyset$, then $ \bigcup_{\alpha \in A} E_{\alpha}$ is connected.
$\textbf{My Attempt:}$ I am reasonably certain this is not the most succinct proof. However, is it valid?
Suppose on the contrary that $\bigcup_{\alpha \in A} E_{\alpha}$ is not connected. Then there exists open, nonempty sets $U$, $V$ such that $$U \cup V = \bigcup_{\alpha \in A} E_{\alpha}$$ and $U \cap V = \emptyset$. Since $\bigcap_{\alpha \in A} E_{\alpha} \neq \emptyset$, there exists at least one element $x$ such that $ x \in \bigcap_{\alpha \in A} E_{\alpha} \neq \emptyset$. Then $x \in U$ or $x \in V$.
Assume WLOG that $x \in U$. Since $V$ is non-empty there exist at least on $E_{\tilde{\alpha}} \in \bigcup_{\alpha \in A} E_{\alpha}$ such that there exists $y \in E_{\tilde{\alpha}}$ and $y \in V$. From above, we must also have $x \in E_{\tilde{\alpha}}$. Let $$M := \{ u \in U : u \in E_{\tilde{\alpha}} \}$$ and $$N:= \{ v \in V : v \in E_{\tilde{\alpha}} \}$$ Then since $U$ and $V$ are open, for each $u \in M$, $v \in N$ we have that there exists an open set $W_u \subset U$ and $R_v \subset V$ where $W_u$, $R_v$ are open. Let $$X:= \bigcup_{u \in M} W_u \bigcap E_{\tilde{\alpha}}$$ $$Y:= \bigcup_{v \in N} R_v \bigcap E_{\tilde{\alpha}}$$ Then $X$ and $Y$ both open with respect to the relative topology on $E_{\tilde{\alpha}}$, $X \cup Y = E_{\tilde{\alpha}}$ and $X \cap Y= \emptyset$. This implies that $E_{\tilde{\alpha}}$ disconnected and thus contradicts that each $E_{\alpha}$ is a connected subset. Thus $ \bigcup_{\alpha \in A} E_{\alpha}$ is connected.
