Any two disjoint open sets are the interior and exterior of some set
Here is the topological property that should suffice to make this statement true:
A topological space $X$ is resolvable iff there is a set $T\subseteq X$ such that $T$ and $T^c$ are both dense.
Then $\Bbb R^2$ is resolved by $\Bbb Q^2$.
First, we prove the theorem for regular open sets $A,B$ (note that a regular open set $U$ is one such that $\text{int }\bar U=U$). Now let $X$ be resolved by $T$, and let $A,B$ be disjoint regular open sets in $X$, and let $S=A\cup (T\setminus B)$. I claim that $\DeclareMathOperator{\ntr}{int}\ntr S=A$ and $\DeclareMathOperator{\ext}{ext}\ext S=B$. Since the interior function respects set inclusion, we have:
$$A=\ntr A\subseteq\ntr A\cup (T\setminus B)=\ntr S$$ $$B=\ntr B=\ntr A^c\cap B\subseteq\ntr A^c\cap(T^c\cup B)=\ext S$$
For the converse, note that $(\ntr\bar A\cup T)\setminus\bar A\subseteq (\bar A\cup T)\setminus\bar A\subseteq T$, and $(\ntr\bar A\cup T)\setminus\bar A=(\ntr\bar A\cup T)\cap\ext A$ is open, so $(\ntr\bar A\cup T)\setminus\bar A\subseteq\ntr T=\emptyset$. Therefore $\ntr\bar A\cup T\subseteq\bar A$, so $$\ntr S\subseteq\ntr A\cup T\subseteq\ntr\bar A\cup T=\ntr\ntr\bar A\cup T\subseteq\ntr\bar A=A$$ (since $A$ is regular open).
Similarly for $B$, $\ext S\subseteq\ntr T^c\cup\bar B\subseteq\ntr\bar B=B$ because $(\ntr T^c\cup\bar B)\setminus\bar B$ is an open set contained in $T^c$, which has empty interior because $T$ is dense.
Using the above as a lemma, we can generalize the theorem to any pair of open sets $A,B$. Let $\DeclareMathOperator{\irr}{irr}\irr U=\ntr\bar U\setminus U$. Then $U$ is regular iff $\irr U=\emptyset$ (which is to say, $\irr U$ is the set of "irregular points" of $U$). Now for arbitrary open disjoint $A,B$, let $Y=X\setminus(\irr A\cup \irr B)$ as a subspace. Then $\irr_YA=\irr_YB=\emptyset$, so there is a set $Q\subseteq Y$ such that $\ntr_Y Q=A\cap Y=A$ and $\ext_Y Q=B\cap Y=B$. Note that $\irr A$ is disjoint from $B$ and vice-versa, since $\irr A\subseteq\bar A$, and $B=\ntr B\subseteq \ntr A^c=\bar A^c$.
I claim that for $S=Q\cup\irr B$, $\ntr S=A$ and $\ext S=B$. Now $\ntr S$ is disjoint from $\irr B$ because $\ntr S$ and $B$ are disjoint (for the same reason that $A$ and $\irr B$ are disjoint), so $\ntr S\subseteq Y$ is open, and thus $\ntr S\subseteq\ntr_Y Q=A$. Similarly, $\ext S\subseteq B$. Conversely, $A\subseteq Q\subseteq S$, and since $A$ is open, we have $A\subseteq\ntr S$, and similarly $B\subseteq\ext S$. (Details of this paragraph borrowed from Karl's answer)