The closure of a set $X$ is equal to the union of $X$ and its boundary $\partial X$ (those points $b$ in the topological space such that every open set containing $b$ contains at least one point in $X$ and one not in $X$). Let $x \in \overline{A} - \overline{B}$. There are two cases to consider:
- $x \in A$ (and $x \notin \overline{B}$)
- $x \notin A$ and $x \in \partial A$ (and $x \notin \overline{B}$)
You mentioned that case 1 is obvious, so I'll explain case 2. To show that $x \in \overline{A - B}$, assume that $x \notin A - B$ (since then it would also be in $\overline{A-B}$). We'll show that this implies $x \in\partial(A - B)$. To that end, let $U$ be any open set containing $x$. We must show that $U \cap (A-B) \neq \emptyset$. Since $x \notin \overline{B}$, there is an open set $V$ containing $x$ such that $V \cap B = \emptyset$. Now $V \cap U \neq \emptyset$ is an open set containing $x$, so using the fact that $x \in \partial A$, there exists $y \in (V\cap U) \cap A$. This puts $y \in U \cap (A-B)$. Thus, by definition, $x \in \overline{A - B}$, as required.
Hope this helps!