My first instinct with this proof is to assume the opposite of the hypothesis, as in a proof by contradiction.
My work is as follows:
Suppose $(A-B) \cap (B-A) \neq \emptyset$.
Consider an $x \in (A-B) \cap (B-A)$.
If $x \in (A-B) \land x \in (B-A)$
$(x \in A \land x \notin B) \land (x \in B \land x \notin A)$
$(x \in A \land x\notin A) \land (x \in B \land x \notin B)$
These are contradictions. Hence such an element $x$ does not exist.
$\implies (A-B) \cap (B-A) = \emptyset$.
My issue is that it is apparent that assuming that the set is not empty is apparently not good practice? Is there a better way to approach this proof?