Let $A$ and $B$ be subsets of a topological space $X$. If $A$ and $B$ are both open or both closed, then the sets $A -B$ and $B-A$ are separated sets.
My thinking is :
Suppose that $A-B$ and $B-A$ are not separated sets. Let $x\in (A-B)$, then $x\in (B-A)$ since $(A-B)\cap (B-A) \neq \emptyset$. Then $x\in A, x\not\in B$ for $x\in (A-B)$ and $x\not\in A$,$x\in B$ for $x\in (B-A)$, which is contradiction. So, $(A-B)\cap$$（B-A) =$ $\emptyset$.
but how can I prove that $(A-B)\cap \overline{(B-A)}$ = $(A-B)\cap (B-A)$.
What you managed to prove is that the sets $A-B$ and $B-A$ are disjoint. This is not a topological property it belongs to elementary set theory.
In order to prove $(A-B)\cap\overline{(B-A)}=\emptyset$, note that we have two cases: (1) both $A$ and $B$ are open and (b) both are closed. I'll show you how to solve the former, the latter should be similar.
Note that $A-B\subseteq A$ and that $A\cap(B-A)=\emptyset$. From this follows that $A-B\cap \mathscr{C}A=\emptyset$ and $B-A\subseteq\mathscr{C}A$. Since $\mathscr{C}A$ is closed, $\overline{B-A}\subseteq \mathscr{C}A$. Since $A-B\cap\mathscr{C}A=\emptyset$, a fortiori, $A-B\cap \overline{(B-A)}=\emptyset$.