Difference of closures and closure of difference

If $x\notin\operatorname{cl}(A\setminus B)$, then $x$ has an open nbhd $U$ such that $U\cap(A\setminus B)=\varnothing$. And $$U\cap(A\setminus B)=(U\cap A)\setminus(U\cap B)\;,$$ so $U\cap A\subseteq U\cap B$. If $x\notin\operatorname{cl}B$, then $x$ has an open nbhd $V$ such that $V\cap B=\varnothing$. Let $W=U\cap V$. Then

$$W\cap A=W\cap(U\cap A)\subseteq W\cap(U\cap B)=W\cap B=\varnothing\;,$$

so $W\cap A=\varnothing$, and $x\notin\operatorname{cl}A$. Thus, if $x\in(\operatorname{cl}A)\setminus\operatorname{cl}B$, then $x\in\operatorname{cl}(A\setminus B)$.

It is true actually, here is a proof: let $x \in \overline{A} - \overline{B}$ and assume for the sake of contradiction that $x \notin \overline{A-B}$. This last condition is equivalent to "there exists an open set $U$ such that $x \in U$ and $U \cap (A - B) = \emptyset $". Since $x \in \overline{A} - \overline{B} \subset \overline{A}$ then $U \cap A \neq \emptyset$. This implies that $U \subset B$ but then we have a contradiction since $x \in U \subset B$ and $x \in \overline{A} - \overline{B}$.

What is false is the other inclusion: take $A = (0,2)$ and $B = (0,1)$. Then $\overline{A - B} = [1, 2]$ which is not a subset of $\overline{A} - \overline{B} = (1,2]$.

I hope it helps :)