Closure in a topological space Let $(X, \tau)$ be a topological space. Let $A,B$ be subsets of $X$. Show that $cl(A \cup B)$ $\subset$ $cl(A)$ $\cup$ $cl(B)$
Proof:
Let x be in the closure of $A \cup B$. That means for every open set U containing x, $U \cap (A \cup B)$ is non empty. That means for every open set U containing x either $U \cap A$ is non empty or for every open set U containing x $U \cap B$ is non empty. Which means that x$\in cl(A) \cup cl(B)$
Is this proof correct?
 A: Not really convincing to be honest. Your proof fully depends on $U$. In order to prove that $x\in cl(A)$ you need to show that for every open set $U$ which contains $x$ we have that $U\cap A$ is not empty. Same thing about $cl(B)$. What you showed is that for every open neighborhood $U$ of $x$ one of the sets $U\cap A, U\cap B$ is not empty, but you don't know which one of them. So there might be an open neighborhood $U_1$ for which $U_1\cap A$ is not empty and another open neighborhood $U_2$ for which $U_2\cap B$ is not empty. So how exactly do you conclude that $x$ is in the closure of $A$ or $B$? 
I would do it like this: suppose $x\notin cl(A)\cup cl(B)$. Then there are open neighborhoods $U,V$ of $x$ such that $U\cap A,V\cap B$ are both empty. But then $U\cap V$ is an open neighborhood of $x$ and its intersection with $A\cup B$ is empty. This is a contradiction to $x\in cl(A\cup B)$. 
A: I think not, suppose there existed $U,V$ open sets containing x so that $U \cap A=\emptyset$(but $U \cap B\neq\emptyset$) and $V \cap B=\emptyset$(but $V \cap A\neq\emptyset$).
Then $x\not\in cl(A)$ and  $x\not\in cl(B)$.
Now of course, what you claimed is true, hopefully you can modify your argument to address this.
Edit: However, your proof in Interior topology was correct, then noticying that $cl(X-A)=X-int(A)$ you can build a proof for what you want
A: No, it is not correct. You know that every open set $U$ to which $x$ belongs intersects $A\cup B$. It is not obvious just from this that $U$ intersects $A$ and $B$.
Note that if $x\notin\overline B$, then, since $\overline B$ is a closed set, then $\overline B^\complement$ is an open set to which $x$ belongs. Therefore, $U\cap\overline B^\complement$ intersects $A\cup B$, for every open set $U$ to which $x$ belongs. But it does not intersect $B$. So, it intersects $A$. So, $x\in\overline A$.
A: Not correct (see the other answers).
As a sort of continuation of my answer on your former question now sortlike characterizations of $\mathsf{cl}(C)$:

The closure of set $C$ is the intersection of all closed sets that contain $C$ as a subset.

or:

The closure of set $C$ is the smallest closed set that contains $C$ as a subset.

Now observe that $\mathsf{cl}(A)\cup\mathsf{cl}(B)$ is evidently a closed set that contains $A\cup B$ as a subset, so we may conclude immediately that: $\mathsf{cl}(A\cup B)\subseteq\mathsf{cl}(A)\cup\mathsf{cl}(B)$
A: As noted bellow it is not correct. It can be that some open sets intersect A but not B and other intersect B but not A.  You can build a similar arguement to the one you made, but based on $x  \notin cl(A) \cup cl(B) \Rightarrow x\notin cl(A \cup B)$. It has been shown here: Proof of $cl(A \cup B)=cl(A) \cup cl(B)$ in the second answer (not the one that was accepted).
In fact it is:
1. $cl(A \cup B) = cl(A) \cup cl(B)$


*$int(A \cap B) = int(A) \cap int(B)$

*$int(A) \cup int(B) \subseteq int(A \cup B)$

*$cl(A) \cap cl (B) \subseteq cl(A \cap B)$
You can see the proofs here:
https://www.youtube.com/watch?v=T1GXwknimMg&list=PL-_cKNuVAYAX_LKTQPzvg5lKaOI8_EpjL&index=3 
starting at 25:30.
A: The correct conclusion from $x\in\operatorname{cl}(A\cup B)$ is

for every neighborhood $U$ of $x$, then $U\cap(A\cup B)\ne\emptyset$

that translates into

for every neighborhood $U$ of $x$, then either $U\cap A\ne\emptyset$ or $U\cap B\ne\emptyset$.

Symbolically, denoting by $\mathscr{U}_x$ the family of neighborhoods of $x$:

(1) $\forall U \bigl( U\in\mathscr{U}_x \to (U\cap A\ne\emptyset)\lor(U\cap B\ne\emptyset) \bigr)$

You are stating, instead,

(2) $\bigl(\forall U (U\in\mathscr{U}_x \to U\cap A\ne\emptyset\bigr) \lor \bigl(\forall U (U\in\mathscr{U}_x \to U\cap B\ne\emptyset\bigr)$

Can you see the difference between (1) and (2)?
Correct argument. Suppose $x\notin\operatorname{cl}(B)$. Then there exists $U_0\in\mathscr{U}_x$ such that $U_0\cap B=\emptyset$. If $U\in\mathscr{U}_x$, then $(U\cap U_0)\cap B=\emptyset$; on the other hand, $U\cap U_0\in\mathscr{U}_x$ implies $(U\cap U_0)\cap(A\cup B)\ne\emptyset$. Therefore $\emptyset\ne(U\cap U_0)\cap A\subseteq U\cap A$. This proves $x\in\operatorname{cl}(A)$.
Note that (2) is actually true in this particular context, because of the property that $\mathscr{U}_x$ is closed under intersections.
