Proof with intersection of closed A set and interior of B set I'm stuck at this proof.
Let $\mathrm{A}$ and $ \mathrm{B}$ be disjoint sets in topological space $ (\mathrm{X}, \tau) $. Prove that if $\mathrm{A}$ is open set in $\mathrm{X}$ then $ \overline{\mathrm{A}} \cap \operatorname{Int}{\overline{B}} = \emptyset$.
Let's suppose that $ \overline{\mathrm{A}} \cap \operatorname{Int}{\overline{B}} \neq\emptyset$.
Hence exists $\mathrm{U}$ - neighbourhood of $x$ such that $ x \in U \subset \overline{\mathrm{B}}$.
By definition of closedness for $\overline{\mathrm{A}}$ comes $ \mathrm{U} \cap \mathrm{A} \neq \emptyset$. So there exists open set, that is completely included in $\overline{\mathrm{B}}$, but its intersection with $\overline{\mathrm{A}}$ is not empty.
I have no idea what would be next. Could you give me a tip?
 A: $A$ is open and $A\cap B=\varnothing$, so $A\cap\operatorname{cl}B=\varnothing$, and therefore $A\cap\operatorname{int}\operatorname{cl}B=\varnothing$. Let $U=\operatorname{int}\operatorname{cl}B$; $U$ is open, and $U\cap A=\varnothing$, so ... ?
A: If $A$ is open, then $X\smallsetminus A$ is closed, and hence $B\subset X\smallsetminus A$, implies that $\overline{B}\subset \overline{X\smallsetminus A}= X\smallsetminus A$, and that
$$
\mathrm{Int}\,(\overline{B})\subset \mathrm{Int}\, (X\smallsetminus A)=X\smallsetminus \overline{A},
$$
and thus
$$
\mathrm{Int}(\overline{B})\cap \overline{A}=\varnothing.
$$
Note. We have used the fact that: $\,\,\mathrm{Int}\, (X\smallsetminus A)=X\smallsetminus \overline{A}$.
A: First note that $A\cap\overline{B}$ is empty. To see that, assume $x\in A\cap\overline{B}$, so $A$ is a neighborhood of $x$, thus intersects $B$ - contradiction. 
Suppose now $x\in\overline{A}\cap Int\overline{B}$, so there as an open $U$ satisfying $x\in U\subset\overline{B}$. Since $x\in\overline{A}$, $U$ intersects $A$, thus so does $\overline{B}$, in contradiction to the previous paragraph.
