Are co-semidecidable languages separated by decidable ones?

Can I use principles of set theory in order to solve the following question?


For every $A,B \in \mathsf{co\text{-}RE}$ with $A \cap B = \emptyset$, there is a separating language $C$ with $A \subseteq C$ and $B \cap C = \emptyset$ so that $C$ is recursive.

This is a classical exercise in computability.
Since $A,B\in coRE$, there exist TMs $M,N$ that recognize $\overline{A}$ and $\overline{B}$ respectively.

Consider the following TM $K$:
Given a word $w$, run in parallel $M$ and $N$ on $w$. That is, simulate $M$ for a single step, and then $N$ for a single step, and repeat.

If $M$ accepts at any point, reject.
If $N$ accepts at any point, accept.
(if both accept, reject, so we first check $M$ for acceptance).

Define $C=L(K)$.

If $A\cap B=\emptyset$, then $\overline{A}\cup\overline{B}=\Sigma^*$, so either $M$ or $N$ eventually accepts, so $K$ always halts. Thus, $C$ is decidable.

We claim that $C$ satisfies the requirements: if $x\in A$, then $M$ never accepts $x$, so $N$ accepts $x$, so $K$ accepts, so $x\in C$. Thus, $A\subseteq C$.

If, by way of contradiction, $x\in B\cap C$, then $K$ accepts $x$, which means $N$ accepts $x$, so $x\notin B$ - contradiction. So $B\cap C=\emptyset$.

We conclude that $C$ satisfies the requirements.

StackExchange Computer Science Q#9843, answer score: 9