TITLE: Is this proof of $\overline{\cup}A_\alpha \subset \cup \overline{A_\alpha}$ valid?
QUESTION [2 upvotes]: This is a problem from my topology homework:  
Let $x \in \overline{\cup}A_\alpha$.  Then for every $U(x)$, $U(x) \cap (\cup A_\alpha) \neq \emptyset \rightarrow U(x) \cap A_\alpha \neq \emptyset$ for some $\alpha \rightarrow x \in \overline{A_\alpha} \rightarrow x \in \cup \overline{A_\alpha}$.
Is the proof valid?  If not, where is the mistake?
As far as I can tell, this proof seems to work. However, the "if not" leads me to believe it's missing at least a small step that I'm overlooking.  
I proved the other direction similarly:
Let $x \in \cup \overline{A_\alpha}$.
This implies that $x$ is in at least one of the $A_\alpha$'s, i.e. $\exists \alpha$ such that $x \in \overline{A_\alpha}$. By the definition of closure, this means that $U(x) \cap A_\alpha \neq \emptyset$.  Thus $x \in A_\alpha$, so we may say $x \in \cup A_\alpha$. So  $U(x) \cap (\cup A_\alpha) \neq \emptyset$.  Thus, by the definition of closure, $x \in \overline{\cup}A_\alpha$. 
I'm not entirely confident in topology so any feedback or help would be greatly appreciated.  Thanks!

REPLY [1 votes]: There is a small mistake,
$U(x) \cap A_{\alpha}$ for some $\alpha$ does not imply that $x\in \overline{A_{\alpha}}$.
On the second part, there is also a mistake.
$U(x)\cap A_{\alpha}\not=\emptyset$ doesnot imply that $x\in A_{\alpha}$.
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