Any errors? My proof that if $\left(A - B\right) \cup \left(B - A\right) = A \cup B$, then $A \cap B = \emptyset$

Question

If $A\cup B=(A-B)\cup(B-A)$, then $A\cap B=\emptyset$.

Proof by Contrapositive. If $A\cap B\ne\emptyset$, then $A\cup B\ne(A-B)\cup(B-A)$. Suppose that there exists a member of $A\cap B$, $x$. Then, $x\notin(A-B)$ because $x$ is in $B$. Similarly, $x\notin(B-A)$ because $x$ is in $A$. So, $x\notin(A-B)\cup(B-A)$. However, since $x$ belongs to $A$, $x\in A\cup B$. Therefore, $A\cup B\not\subseteq(A-B)\cup(B-A)$, and $A\cup B\ne(A-B)\cup(B-A)$.

Are there any errors or room for improvement in the above proof?

Answer 1

$(A - B)\cup(B - A) = (A \cap B^c) \cup (B \cap A^c) =$
$(A \cup B) \cap (A^c \cup B^c) = (A \cup B) - (A \cap B)$
$= A \cup B$
provides a direct proof.