Sorry for long title - the question is a bit unwieldy. To state the question precisely, I'm wondering about the following proposition:
Let $\Sigma = \{0,1\}$. If $A$ and $B$ are Turing-recognizable and $A \cup B = \Sigma^*$, then there exists a Turing-decidable language $C$ such that $A \cap \overline{B} \subseteq C$ and $\overline{A} \cap B \subseteq \overline{C}$ (equivalently, $A - B \subseteq C$ and $B - A \subseteq \overline{C}$
So, my current train of thought is as follows. Because $A \cup B$ is $\Sigma^*$, we have that $\Sigma^*$ is partitioned into exactly three disjoint sets: $A - B$, $A \cap B$, and $B - A$. That is, given $x \in \Sigma^*$, $x$ is in exactly one of these sets.
The basic way I can think of to obtain a decidable language is to enumerate one in lexicographic order (which always yields a decidable language). So one idea would be to enumerate $\Sigma^*$ in lexicographic order, and print a string if and only if it is not in $B - A$, but that poses a problem because we would have to decide membership in $B - A$, but they are merely recognizable.
I've thought about enumerating $A$ and $B$ but since they're only recognizable, the enumerations could be in any arbitrary order, making it impossible to compare them to test for membership in $A-B$ or $B-A$.
Finally, I know that $A$ and $B$ (if they are infinite) have infinite decidable subsets, but I don't think that will help.
I'd appreciate some hints if anyone has any.
Edit: Missed a hypothesis that $\Sigma = \{0,1\}$, which I'm not sure is relevant (I wouldn't think so), but it might be.