If $A\cap B=\emptyset$ and $A$ is an open set, then $\overline A\cap \operatorname{Int} (\overline B)=\emptyset$??
Since $A$ is open, $A^c$ is closed. Since $B$ is disjoint from $A$, $B\subset A^c$ and thus $\overline B \subset A^c$. Thus $A\cap \overline B = \emptyset$.
On the other hand, since $\operatorname{int} (\overline B) \subset \overline B$ by definition and is open, $A\cap \overline B = \emptyset$ implies that $A \subset [\operatorname{int} (\overline B)]^c$, which is closed. Thus $\overline A \subset [\operatorname{int} (\overline B)]^c$ by definition of closure. This is the same as $\overline A \cap \operatorname{int} (\overline B) = \emptyset$.