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?

$A$ 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 ... ?
