What is the general method to show that two disjoint languages are not separable? As an example, suppose we have:
$A = \{\langle M \rangle : M ( \langle M \rangle )$ halts and says ACCEPT$\}$
$B = \{\langle M \rangle : M ( \langle M \rangle )$ halts and says REJECT$\}$
If there are any resources that shed light on the matter, it would be helpful if these are linked, but I'm primarily seeking a proof technique for general problems of this nature (specifically this one, if possible).
I do not know of any general way to know if sets are separable.
There are some condition for which we know separations do exists.
$\Pi_1^0$ or co-recursively enumerable disjoint subsets of the natural numbers can always be separated by a computable ($\Delta_1^0$) subsets of the natural numbers.
In general, for Polish spaces, $\mathbf{\Pi}_\xi^0$ have the separation property in the sense that disjoint $\mathbf{\Pi}_\xi^0$ subsets can always be separated by $\mathbf{\Delta}_\xi^0$ subsets. $\mathbf{\Sigma}_1^1$ disjoint subsets can be separated by $\mathbf{\Delta}_1^1$ sets. $\mathbf{\Pi}_2^1$ disjoint sets can be separated by $\mathbf{\Delta}_2^1$ sets. Higher separations require set theoretic assumptions. For example $V = L$ and determinacy give differing separation property for the higher projective pointclasses.
However, when you are outside of these cases whether or not a separation exists is more of a cases by cases argument. Usually you assume there is a separation and by some diagonalization argument obtain a contradiction.
Suppose there is a computable set $C$ such that $B \subset C$ and $C \cap A = \emptyset$. $C$ being computable means there is a Turing machine $U$ which halts on all inputs and $n \in C$ if and only if $U(n)$ accepts. Now lets ask where is $\langle U \rangle$. If $\langle U \rangle \in B \subseteq C$, then by defintion of $B$, $U(\langle U \rangle)$ halts and rejects. But $\langle U \rangle \in C$ which by defintion of $U$, one must have $U(\langle U \rangle)$ halts and accepts. Contradiction. Suppose $\langle U \rangle \in A$. Then $\langle U \rangle \notin C$. So $U (\langle U \rangle)$ halts and rejects. This implies that $\langle U \rangle \in B \subseteq C$ which is a contradiction again.