The Galois connection between topological closure and topological interior [Update: I changed the question so that $-$ is only applied to closed sets and $\circ$ is only applied to open sets.]
Let $X$ be a topological space with open sets $\mathcal{O}\subseteq 2^X$ and closed sets $\mathcal{C}\subseteq 2^X$. Consider the pair of maps $-:\mathcal{O}\leftrightarrows\mathcal{C}:\circ$ where $-$ is the topological closure and $\circ$ is the topological interior. Under what conditions will it be true that for all $A\in\mathcal{O}$ and $B\in\mathcal{C}$ we have $$A\subseteq B^\circ \Longleftrightarrow A^-\subseteq B \,\,?$$
[Update: Darij showed that this condition holds for all topological spaces.] If this condition does hold then by general nonsense (the theory of abstract Galois connections), we obtain an isomorphism of lattices $$-:P\approx Q:\circ$$
where $P$ is the the lattice of sets $A\in\mathcal{O}$ such that $(A^-)^\circ=A$ and $Q$ is the lattice of sets $B\in\mathcal{C}$ such that $(B^\circ)^-=B$, where both $P$ and $Q$ are partially ordered by inclusion.
The existence of this lattice isomorphism makes me wonder: is there a nice characterization of the elements of $P$ and $Q$? Certainly not every open set is in $P$. For example, if $X=\mathbb{R}$ with the usual topology then the set $(0,1)\cup (1,2)$ is open, but $$(((0,1)\cup(1,2))^-)^\circ = ([0,2])^\circ = (0,2) \supsetneq (0,1)\cup(1,2).$$
[Update: I found the answer. See below.]

Put $A=B=S$, where $S$ is any subset of $X$. Then
$$
\overline S\subseteq S \iff S\subseteq\mathring S
$$
hence
$$
S \text{ is closed} \iff S \text{ is open}
$$
Conversely, assume that every open set is closed (which implies that any closed set is open). If $A$ and $B$ are subsets of $X$ such that $\overline A ⊆ B$, then we also have $A ⊆ \overline A ⊆ \mathring B$, and if $A ⊆ \mathring B$, then $\overline A ⊆ \mathring B ⊆ B$. Hence such a space satisfies
$$
\overline A ⊆ B \iff A \subseteq \mathring B
$$
