Closure of the union = Union of closures I have seen that $\text{cl}(A\cup B)=\text{cl}(A)\cup \text{cl}(B)$. However I don't see why this is true. I can see the the right to left inclusion, but I can't see the inclusion from left to right. Say that I have an element $x$  contained only in two open sets one that intersects $A\cup B$ only in $A\setminus B$ and and another that intersects only in $B\setminus A$ isn't this a contradiction?
Edit: I have seen the proof but I still can't understand what is wrong with the counterexaple above 
Thanks in advance

(1) ($\supset$) ::
\begin{align*}
A \subset A \cup B \implies \text{cl}(A) \subset \text{cl}(A \cup B)
\\B \subset A \cup B \implies \text{cl}(B) \subset \text{cl}(A \cup B)
\end{align*}
therefore yielding that $\text{cl}(A) \cup \text{cl}(B) \subset\text{cl}(A \cup B)$
(2) ($\subset$) ::
The subset $\text{cl}(A) \cup \text{cl}(B)$ is closed and contains both $A$ and $B$, therefore $A \cup B \subset \text{cl}(A) \cup \text{cl}(B)$. $\text{cl}(A \cup B)$ is defined to be smallest closed set which contains $A \cup B$, so that any closed set which contains $A\cup B$ also contains $\text{cl}(A \cup B)$. Therefore $\text{cl}(A \cup B) \subset \text{cl}(A) \cup \text{cl}(B)$.
