$A$ is representable as a difference of two closed sets if and only if $\overline A \setminus A$ is closed. 
Show that the set $A$ is representable as a difference of two closed sets if and only if $\overline A \setminus A$ is closed.


My try:
$(\Leftarrow)$ Suppose $\overline A \setminus A$ is closed. Since $\overline A$ is closed and  $A = \overline A \setminus (\overline A \setminus A)$, we see that $A$ is represented as a difference of two closed sets. $\square$
$(\Rightarrow)$ Suppose $A = B\setminus C$ where $B$ and $C$ are closed sets. We need to show that $\overline A \setminus A$ is closed. Since $\overline A = \overline {B\setminus C}$, we have $$\overline A \setminus A = (\overline {B\setminus C}) \setminus (B \setminus C) = \partial (B\setminus C)$$ Since the boundary is closed (?), $\overline A \setminus A$ is closed $\square$

Last arguments of mine seem to be vague not clear. How to solve the problem rigorously?
 A: Proof of $\implies$:
$\overline A \setminus A =\overline  {(B\setminus C)} \cap (B^{c} \cup C)$. Since $\overline  {(B\setminus C)} \cap B^{c} $ is empty we get $\overline A \setminus A =\overline  {(B\setminus C)} \cap  C$ which is closed.
A: The first half is of course fine. The problem with the second half is that
$$\operatorname{cl}(B\setminus C)\setminus(B\setminus C)$$
is not necessarily the boundary of $B\setminus C$. In the real line, for instance, take $B=[0,2]$ and $C=[0,1]$; then $B\setminus C=(1,2]$ and $\operatorname{cl}(B\setminus C)=[1,2]$, so
$$\operatorname{cl}(B\setminus C)\setminus(B\setminus C)=\{1\}\,,$$
but $\operatorname{bdry}(B\setminus C)=\{1,2\}$.
Clearly $A\subseteq B$, and $B$ is closed, so $\operatorname{cl}A\subseteq B$, and therefore
$$(\operatorname{cl}A)\setminus C\subseteq B\setminus C=A\,.$$
On the other hand, $A\cap C=\varnothing$, so $A\subseteq(\operatorname{cl}A)\setminus C$. Thus, $A=(\operatorname{cl}A)\setminus C$.

*

*To complete the argument, show that $(\operatorname{cl}A)\setminus A=C\cap\operatorname{cl}A$, which is of course closed.

A: Partial answer
Your equality $$(\overline {B\setminus C}) \setminus (B \setminus C) = \partial (B\setminus C)$$ is wrong.
For a set $S$ you have
$$\partial S = \overline S \setminus \mathring{S}.$$
