How to show these sets are equal: $\overline{X \setminus A} = X \setminus A^\circ$

Question

Given a metric space $(X,d)$ with $A\subseteq X $, show that $\overline{X \setminus A} = X \setminus A^\circ$

I know that I have to show that $\overline{X \setminus A} \subseteq X \setminus A^\circ$ and that $X \setminus A^\circ \subseteq \overline{X \setminus A}$.

Already I have: since $A^\circ \subseteq A$, $X \setminus A \subseteq X \setminus A^\circ$ and that since $A^\circ$ is open, $X \setminus A^\circ$ is closed. Using the fact that $\overline{X\setminus A}$ is the smallest closed subset to contain $X \setminus A$, $$\overline{X \setminus A} \subseteq X \setminus A^\circ$$ How do I go about proving the other way?

Answer 1

$X\setminus(\overline{X\setminus A})$ is an open set contained in $A$ (hence contained in $A^{\circ}$).

This leads to: $$X\setminus A^{\circ}\subseteq\overline{X\setminus A}$$

Answer 2

Can't you start from the fact that $X \setminus A^\circ$ is closed, thus $X \setminus A^\circ = \overline{X \setminus A^\circ}$