# Do closure and intersection of open sets commute when composed with the interior?

In the Wikipedia article on the interior operator it is stated that on a complete metric space $$X$$ one has for every sequence of open sets $$(A_i)_{i\in \mathbb N}\subset X$$ the relation $$\mathrm{Int}\Big(\bigcap_{i}\overline A_i\Big)=\mathrm{Int}\Big(\overline {\bigcap_{i}A_i}\Big).$$ Now I wonder whether the finite version of this relation, i.e., for any two open sets $$A,B\subset X$$ to have $$\mathrm{Int}\big(\overline A\cap \overline B\big)=\mathrm{Int}\big(\overline {A\cap B}\big),$$ does actually hold in any topological space $$X$$. If so, it should be found in some elementary textbooks, perhaps as an exercise. I searched for a while but it turned out that searching for relations like this is extremely inconvenient - one obtains hundreds of results which are similar but different from the wanted relation. On the other hand, I guess that (if the relation holds in general) there should be an easy direct proof based on playing around with the operations interior, closure, intersection, union, and complement. Since I don't have enough time for this game I'm asking directly for a reference.

Yes, it's a standard fact used in the theory of so-called regular-open sets (sets such that $$\operatorname{int}(\overline{U})=U$$; your sets of the form "interior of closure" are regular open..) and forms part of the proof that those sets form a Boolean algebra under the operations $$U \land V = U \cap V$$, $$U \lor V = \operatorname{int}(\overline{U \cup V})$$ and $$\lnot U = \operatorname{int}U^\complement$$. I think Halmos proves it in his Boolean algebra book, and it's also in Engelking somewhere, I'm sure.
As $$\overline{A \cap B} \subseteq \overline{A} \cap \overline{B}$$, because $$A \subseteq \overline{A}$$ and likewise for $$B$$, so $$\overline{A} \cap \overline{B}$$ is a closed superset of $$A \cap B$$, so it contains $$\overline{A \cap B}$$ too. Taking interior on both sides gives the easy inclusion (we don't need anything on $$A$$,$$B$$):
$$\operatorname{int}(\overline{A \cap B})\subseteq \operatorname{int}(\overline{A} \cap \overline{B})$$
For the reverse inclusion we can use a small lemma that for an open $$U$$ and any $$A$$ we have $$\overline{U \cap A} = \overline{U \cap \overline{A}}$$. Think about it.