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
 A: (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 $\text{cl}(A \cup B) \subset \text{cl}(A) \cup \text{cl}(B)$.
A: Let's use the following definition of closure:

Let $A$ be a subset of $(X,\tau)$. Then the set $A \cup A'$, consisting of the set $A$ and all it's limit points it's called the closure of $A$ and is denoted by $\overline A$.

So, we want to prove that $\overline{(A_1 \cup A_2)} = \overline A_1 \cup \overline A_2$.
Part 1
Let's start by proving that $\overline{(A_1 \cup A_2)} \subseteq \overline A_1 \cup \overline A_2$.
Let $x \in  \overline{(A_1 \cup A_2)}$, Then we have that $x \in (A_1 \cup A_2) \cup (A_1 \cup A_2)'$ (this is the definition of closure).
If $x \in (A_1 \cup A_2)$, then, because $\overline A_1 \cup \overline A_2 = (A_1 \cup A_2) \cup (A_1' \cup A_2') $, we have that $x \in \overline A_1 \cup \overline A_2$.
If $x \in (A_1 \cup A_2)'$, then we have that $x$ is a limit point of the set $A_1 \cup A_2$. By the definition of limit point this means that, for every open set $B \in \tau$ such that $x \in B$, $\exists p \in (A_1 \cup A_2): p \in B$ and $p \neq x$. because $p \in A_1 \cup A_2$ we have that $(p \in A_1) \vee (p \in A_2)$ whitch is the same as saying that $x$ is a limit point of $A_1$ $\vee$ $x$ is a limit point of $A_2$. Thus $x \in (A_1' \cup A_2') \to x \in \overline A_1 \cup \overline A_2$.
This means that  $\overline{(A_1 \cup A_2)} \subseteq \overline A_1 \cup \overline A_2$.
Part 2
Now let's prove that $\overline{(A_1 \cup A_2)} \supseteq \overline A_1 \cup \overline A_2$
Let $x \in \overline A_1 \cup \overline A_2$. Then $x \in (A_1 \cup A_2) \cup (A_1' \cup A_2')$.
If $x \in (A_1 \cup A_2)$, it's trivial that $x \in (A_1 \cup A_2) \cup (A_1 \cup A_2)' = \overline{(A_1 \cup A_2)}$
If $x \in (A_1' \cup A_2')$, then $x$ is a limit point of $A_1$ or a limit point of $A_2$. This tells us that, for every $B \in \tau$ such that $x \in B$, $\exists p \in A_1 \vee \exists k \in A_2: p \in B \vee k \in B$, such that both $p$ and $k$ are different from $x$. But $p \in A_1 \subseteq A_1 \cup A_2$, and $k \in A_2 \subseteq A_1 \cup A_2$. So we have that $p,k \in A_1 \cup A_2$. So $x$ is also a limit point of $A_1 \cup A_2 \to x \in \overline{(A_1 \cup A_2)}$. This implies that: $\overline{(A_1 \cup A_2)} \supseteq \overline A_1 \cup \overline A_2$
Part 1 + Part 2 (Conclusion)
From part 1 we deduced that $\overline{(A_1 \cup A_2)} \subseteq \overline A_1 \cup \overline A_2$, and from part 2 $\overline{(A_1 \cup A_2)} \supseteq \overline A_1 \cup \overline A_2$. This leads us to the conclution that $\overline{(A_1 \cup A_2)} = \overline A_1 \cup \overline A_2$
A: Let $x\in Cl(A\cup B)$ then every open set containing $x$ intersects $A\cup B$.  Thus for $x \in U_\alpha$ where $U_\alpha$ is open in $X$.  If $U_\alpha$ intersects $A$, then $x \in Cl(A)$ else $x \in Cl(B)$ either way $x \in Cl(A)\cup Cl(B)$. 
A: We have to proof that $\overline{A \cup B} = \overline{A} \cup \overline{B}$
$(\subset)$
\begin{align}
x \notin \overline{A \cup B} &  \iff \exists V_x (\text{neighborhood of x})  : V_x \cap (A \cup B) = \varnothing\\
&\iff(V_x \cap A) \cup (V_x \cap B) = \varnothing\\
&\implies V_x \cap A= \varnothing \text{ and }  V_x \cap B = \varnothing\\
&\iff x \notin \overline{A} \text{ and } x \notin \overline{A}\\
&\iff x \notin \overline{A} \cup \overline{B}
\end{align}
$(\supset)$
\begin{align}
x \in \overline{A} \cup \overline{B}  & \iff x \in \overline{A} \text{ or } x \in \overline{B}\\
&\iff \forall V_x : V_x \cap A \ne \varnothing \text{ or } V_x \cap B \ne \varnothing\\ 
&\implies (V_x \cap A) \cup (V_x \cap B) \ne  \varnothing\\
&\iff V_x \cap (A \cup B) \neq \varnothing\\
&\iff x \in \overline{A \cup B}
\end{align}
A: In your proposed counterexample, you've forgotten that open sets are closed under finite intersection. 
So if $x$ has a neighborhood that only meets $A\cup B$ in $B\setminus A$ and a neighborhood that only meets $A\cup B$ in $A\setminus B$, then the intersection of these neighborhoods doesn't meet $A\cup B$ at all. 
A: Infinite point sequence from  (A ∪ B) contains an infinite subsequence from A or contains an infinite subsequence from B
A: I got stuck at the same point. I was able to use contraposition to make things easier - I believe this works:
We want to show that $x\in (H\cup K)' \Rightarrow x \in H' \cup K'$ (and thus $x\in \overline{H} \cup \overline{K}$ )
By way of contraposition, suppose $x \notin H' \cup K'$
Thus there exists an open $D$ containing $x$ which contains no points of either $H$ or of $K$ distinct from $x$. In other words, $D$ contains no points of $H\cup K$ distinct from $x$.
Thus $x \notin(H\cup K)'$
