For $U,V \in \tau $ st $U \cap V = \emptyset$ prove $int(cl(U)) \cap int(cl(V)) = \emptyset$ Let $(E, \tau)$ be a topological space,
For $U,V \in \tau $ st $U \cap V = \emptyset$. Prove $int(cl(U)) \cap int(cl(V)) =\emptyset$
I know $cl(A) = \displaystyle  \bigcap_{A \subset S}S$ so $cl(A) \cap cl(B) = \displaystyle  \bigg(\bigcap_{A \subset S}S\bigg) \cap \bigg(\displaystyle  \bigcap_{B \subset T}T\bigg) $ where $S,T \subset E$ are closed
now need $int\displaystyle  \bigg(\bigcap_{A \subset S}S\bigg) \cap int\bigg(\displaystyle  \bigcap_{B \subset T}T\bigg)$
And $\displaystyle int(C) = \bigcup_{O \subset C}O, O$ is an open subset of $C \subset E$ 
So we get $\displaystyle \Bigg(\bigcup_{O \subset \bigcap_{A \subset S}S}O\Bigg) \cap \Bigg(\bigcup_{P \subset \bigcap_{B \subset T}P} P\Bigg)$ , $O \subset cl(A), P \subset cl(B)$ : open subsets
Also, for two sets the interior of the intersection equals the intersection of the the interiors. so
$\displaystyle \Bigg(\bigcup_{O \subset \bigcap_{A \subset S}S}O\Bigg) \cap \Bigg(\bigcup_{P \subset \bigcap_{B \subset T}P} P\Bigg) = \bigcup_{U \subset (\bigcap_{A \subset S}S)\, \cap \,( \bigcap_{B \subset T}T)}U, U$:open  
Not sure how to get this equal to $\emptyset$.
 A: If you use the characterisation of $\operatorname{cl}(A)$ as those $x$ such that every open set that contains $x$, intersects $A$, one could reason as follows:
Suppose $O = \operatorname{int}\operatorname{cl}(U) \cap \operatorname{int}\operatorname{cl}(V)$ is non-empty (it's already open as the intersection of two open sets). Then $O$ is a subset of $\operatorname{cl}(U)$, so $O$ must intersect $U$, and so $O \cap U$ is a non-empty open set as well. As $O \cap U \subseteq O \subseteq \operatorname{cl}(V)$, we have that $O \cap U$ must intersect $V$ as well, but this would imply that $U$ intersects $V$, which is false. So contradiction. 
Added 
If the only definition of closure you know is $\operatorname{cl}(A) = \cap \{ F: F \text{ closed}, A \subseteq F\}$, we can quite easily show that $\operatorname{cl}(A)$ is the set $\{x: \forall O \subseteq X: (O \text{ open and } x \in O) \rightarrow O \cap A \neq \emptyset\}$, the set of so-called adherence points of $A$.
Suppose $x$ is in the first intersection. Take $O$ open and containing $x$. Then $X \setminus O$ is closed, and it does not contain $x$, so it cannot be a closed set containing $A$ ($x$ is by definition in all such sets!) so $A \nsubseteq X\setminus O$, which means exactly that $O$ intersects $A$, as required: $x$ is thus an adherence point of $A$. And, on the other hand, if $x$ is an adherence point, let $F$ be any closed set that contains $A$ as a subset. If $x \notin F$ (striving for a contradiction) then $X \setminus F$ is open and contains $x$ and so must intersect $A$, contradicting $A \subseteq F$, so $x \in F$, and so $x$ is also in the closure of $A$ (i.e. the intersection definition).
This fact should be well-known, but might not have been covered yet in your text. It's one thing to define the closure as the minimal closed set surrounding a set, but it's often better to have an explicit test to see whether some $x$ is in this closure, namely: is it an adherence point of the set.
