Prove that: If $X$ is a topological space and $A$ and $B$ are two subsets of $X$ then,$Cl(A) \cup Cl(B) = Cl(A \cup B) $ Prove that: If $X$ is a topological space and $A$ and $B$ are two subsets of $X$ then,$Cl(A) \cup Cl(B) = Cl(A \cup B) $ where $Cl(H)$ means the closure f the subsets $H $of $X$.
I was able to prove that $Cl(A) \cup Cl(B) \subset Cl(A \cup B) $. I don't know how to prove the other way. The problem is:
If $x\in Cl(A \cup B) $ then every neighborhood $U$ of $x$ intersect $A \cup B$ so every neighborhood $U$ of $x$ intersects either $A$ or $B$ or both. But it may happen that some neighborhoods intersect $A$ but not $B$ and the rest of them intersect $B$ but not $A$, in which case we get $Cl(A \cup B) \not\subset Cl(A) \cup Cl(B)$. How to proceed?
This problem is from Munkres topology. 
 A: Hint: As $A \subseteq \operatorname{Cl}(A)$ and $B \subseteq \operatorname{Cl}(B)$, $A\cup B \subseteq \operatorname{Cl}(A)\cup\operatorname{Cl}(B)$. Now note that $\operatorname{Cl}(A)\cup\operatorname{Cl}(B)$ is closed.
A: So we want to show that $\operatorname{Cl}(A \cup B) \subseteq \operatorname{Cl}(A) \cup \operatorname{Cl}(B)$. There are two ways (at least) to go about this:
Using the definition with neighbourhoods that you use: let $x$ be in $\operatorname{Cl}(A \cup B)$, but suppose $x$ is not in $\operatorname{Cl}(A) \cup \operatorname{Cl}(B)$. So $x \notin \operatorname{Cl}(A)$ means that there is an open neighbourhood $U$ of $x$ that does not intersect $A$, and $x \notin \operatorname{Cl}(B)$ means that there is an open neighbourhood $V$ of $x$ that does not intersect $B$. Then $U \cap V$ is also an open neighbourhood of $x$ and it misses $A$ as the larger $U$ already does, and it misses $B$ because $V$ already does. But then we have an open neighbourhood of $x$ that misses $A \cup B$ and this contradicts that $x \in \operatorname{Cl}(A \cup B)$. So we are done.
Using the fact (theorem) that $\operatorname{Cl}(C)$ (for any $C \subset X$) is the smallest closed set that has $C$ as a subset: $A \subseteq \operatorname{Cl}(A)$ and $B \subseteq \operatorname{Cl}(B)$, so $A \cup B \subseteq \operatorname{Cl}(A) \cup \operatorname{Cl}(B)$. Now, assuming you have proved that the finite union of closed sets is closed, we conclude that $A \cup B$ is thus contained in the closed set $\operatorname{Cl}(A) \cup \operatorname{Cl}(B)$. As $\operatorname{Cl}(A \cup B)$ is the smallest such closed set, we know that $\operatorname{Cl}(A \cup B) \subseteq \operatorname{Cl}(A) \cup \operatorname{Cl}(B)$ and we are done. But this is of course assuming you know that the closure of a set is the smallest subset around that set and closed sets are closed under finite unions. 
A: To adress your concern if $U$ intersects $A$ but not $B$ and $O$ intersects $B$  but not $A$ then $U\cap O$ is open and intersects neither.
A: A very useful lemma here is the fact that $Cl(A)$ is the minimal closed set containing $A$ (Verify for yourself!).
Assume for the sake of contradiction that the two sets are not equal. We find that, just as Michael said, $Cl(A) \cup Cl(B)$ is closed. In addition, we must have $A \cup B \subset Cl(A) \cup Cl(B)$. Combining this with $Cl(A) \cup Cl(B) \subset Cl(A \cup B)$ gives us the desired result since otherwise we arrive at a contradiction due to the lemma. 
