Proposed proof of set theoretic result I am tasked with proving the following:
$$ (A - B)\cap (B-A) = \varnothing $$
My Attempt:
Suppose there exist a $x \in (A - B)\cap (B-A) $ then:
 \begin{align*}
    x \in (A - B)\cap (B-A) &\iff (x \in A \land x \notin B) \land  (x \in B \land x \notin  A) \\
    &\iff  x \in A \land ( x\notin B \land  x  \in B) \land x \notin A \\
    &\iff  x \in A \land ( x\in B \land  x  \notin B) \land x \notin A \\
    &\iff  (x  \in A  \land x \notin A) \land ( x\in B \land  x  \notin B)\\
    &\iff  x \in (A-A) \cap (B-B)
  \end{align*}
Since $B-B = \varnothing $ and $A-A = \varnothing$ 
then $x \in  \varnothing $. An contradiction 
Is this attempt  correct? Can it be improved in anyway?

Your proof is fine.
A shorter proof could go by noting that if there is an $x\in (A-B)\cap (B-A)$ then $x\in A-B$ and $x\in B-A$. The first will imply $x\in A$ while the second $x\not\in A$, so we'll have a contradiction.
