If $\{E_\alpha\}$ is connected, $\bigcap\limits_{\alpha\in A}E \neq \emptyset$, then $\bigcup\limits_{\alpha\in A}E$ is connected If $\{E_\alpha\}_{\alpha\in A}$ is connected in $\mathbb{R}^n$, $\bigcap\limits_{\alpha\in A}E_\alpha \neq \emptyset$, then $\bigcup\limits_{\alpha\in A}E_\alpha$ is connected.
I have zero intuition on how to do a proof of this statement. Please help.
I started it by contradiction like this: suppose that $\bigcup\limits_{\alpha\in A}E_\alpha$ is not connected. Then there exist non-empty disjoint separating sets $U$ and $V$, such that $U$ and $V$ are relatively open in $\bigcup\limits_{\alpha\in A}E_\alpha$, and $U\cup V= \bigcup\limits_{\alpha\in A}E_\alpha$. But I have no idea what to do next.
 A: Hint: Let $y\in \cap E_\alpha \subset \cup E_\alpha$. Then $y\in U$ or $y\in V$. **Fixing each $\alpha$, consider $U_\alpha = U\cap E_\alpha$ and $V_\alpha = V\cap E_\alpha$. They are open, disjoint and $U_\alpha \cup V_\alpha = E_\alpha$. Do you see how to use the connectivity now? 
(Let's say, if $y\in U$. Then $y\in U_\alpha$ is nonempty. Thus $V_\alpha$ has to be empty as $E_\alpha$ is connected).
A: $X$ is connected iff the only nonempty clopen (closed and open) subset of $X$ is all of $X$. 
$E:=\bigcup _{\alpha\in A}E_\alpha$ is connected: Let $U$ be a nonempty clopen subset of $E$. We show $U=E$. There exists $\beta\in A$ such that $U\cap E_\beta\neq\varnothing$. Since $E_\beta$ is connected, we have $E_\beta\subseteq U$. Now let $p\in \bigcap_{\alpha\in A}E_\alpha$. We have $p\in U$. So $U\cap E_\alpha\neq\varnothing$ for each $\alpha\in A$. Since each $E_\alpha$ is connected, we have $E_\alpha\subseteq U$ for each $\alpha\in A$. That is $E\subseteq U$. So $U=E$. Thus $E$ is connected.
