Any two disjoint open sets are the interior and exterior of some set 
For a topological space $X$, given any two open sets $A,B$, there is a set $S\subseteq X$ such that $\DeclareMathOperator{\ntr}{int}\ntr S=A$ and $\DeclareMathOperator{\ext}{ext}\ext S=B$.

Is this true for $X=\Bbb R^2$? If so, what topological properties on $X$ are needed for a general topological space to satisfy this? If not, are there any restrictions that I forgot to add?
 A: 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)
A: Originally, this answer and edits to Mario Carneiro's answer were essentially the same, independently discovered, arguments. I subsequently noticed an equivalent approach that makes better use of my Lemma that I have elucidated in this revision. I refer to the portions of Mario's answer which treat the specific case in which the open sets are regular.
$\DeclareMathOperator{\interior}{int}\DeclareMathOperator{\exterior}{ext}\DeclareMathOperator{\irr}{irr}$For any subset $A\subseteq X$, define $\irr A=\{x\in\interior\bar A\colon x\notin A\}$. Now, suppose $A$ is open in $X$. Then $A\subseteq\interior\bar A$ means that $A\cup\irr A=A\cup(\interior\bar A\setminus A)=\interior\bar A$ is an open subset of $X$. The fact that it is contained in $\bar A$ means that it is regular.
Lemma. For any disjoint open sets $A,B$ in a topological space $X$, the sets $A\cup\irr A$ and $B\cup\irr B$ are also disjoint.
Proof. If $A\cap\irr B$ is nonempty, then $A$ has a nonempty (open) intersection with the interior of $\bar B$. But $A\cap B=\emptyset$, so $\bar B\setminus B$ contains an open set, contradiction. It follows that $A\cap\irr B=\emptyset=\irr A\cap B$. It only remains to show that $\irr A$ is disjoint from $\irr B$. If $x$ is in both, then let $U$ be a neighborhood of $x$ contained in both $\bar A$ and $\bar B$. We have, $$\emptyset\ne U\cap B\subseteq U\cap\interior A^c\subseteq\bar A\cap\interior A^c=\emptyset,$$ a contradiction. q.e.d.
Therefore, if $X$ is a topological space with a dense co-dense subset, then as Mario Carneiro showed there exists $S_0\subset X$ such that $\interior S_0=A\cup\irr A$ and $\exterior S_0=B\cup\irr B$ for any disjoint open sets $A,B\subseteq X$. 
Observe that $A\subseteq\interior(S_0\setminus\irr A)\subseteq(\interior S_0)\setminus\irr A=A$. Define $S=S_0\cup \irr B\setminus\irr A$. If $U$ is an open subset of $S$, then $U\setminus\irr B$ is open in $S_0\setminus\irr A$, so $U\subseteq A$ by our initial calculation. It follows that $\interior S=A$. To see that $\exterior S=B$, just note the following symmetry: $S^c=S_0^c\cup\irr A\setminus\irr B$.
A: 
Let $X$ be a resolvable space with a dense/co-dense subset $D$, and 
  $A, B$ disjoint open subsets of $X$.
  Then the set $S = A \cup (D\setminus \overline{A}\setminus\overline{B}) \cup (\overline{B}\setminus B)$ has $\operatorname{int} S=A$ and $\operatorname{ext} S=B$.

Clearly $A \subset \operatorname{int} S$. To see the reverse inclusion, note
that $S \subset A \cup \operatorname{ext} A \cup \overline{B}$. 
Obviously a point in $\overline{B}$ cannot be interior to any subset of $X\setminus B$.
Because the complement of $D\setminus \overline{A}\setminus\overline{B}$ is dense and the complement of $\overline{B} \setminus {B}$ is dense and open,
their intersection is everywhere dense, so a point in 
$S \cap \operatorname{ext} A$ cannot be interior to $S$. Thus we have shown
that $\operatorname{int} S = A$.
Because $X\setminus\overline{A}\setminus\overline{B}$ is open, we have the inclusion
$$
X\setminus\overline{A}\setminus\overline{B} \subset \overline{X\setminus\overline{A}\setminus\overline{B}} = \overline{D \cap (X\setminus\overline{A}\setminus\overline{B})} = \overline{D \setminus\overline{A}\setminus\overline{B}}
$$
which implies that $X\setminus B \subset \overline{S}$, and because $B$ is
open we have
$$
 \overline{D \setminus\overline{A}\setminus\overline{B}}\subset \overline{D\setminus \overline{B}} \subset \overline{D}\setminus \operatorname{int}B
= X\setminus B
$$
hence $\overline{S} \subset  X\setminus B$. Thus we conclude that $\operatorname{ext} S = B$.
