Let $A$ be a subset of a metric space $(X,d)$. For a subset $B$: let $B^{-}=\cap\{F: F \ \text{is closed and} \ B \subset F \}$,
$\mathrm{Int}B=\cup\{G:G \ \text{is open and} \ G\subset B \}$
and $X-B$ the complement of $B$ in $X$
I wish to prove that $\mathrm{Int}(A)=X-(X-A)^{-}$
My attempt: (first direction ) suppose $y \in X-(X-A)^{-}$ , since $(X-A)^{-}$ is closed in $X$ it follows that the complement is open in $X$ and thus there is an open ball in $X-(X-A)^{-}$ which contains all of $y$, call it $B$.
Question: Can I conclude that $B$ must be a subset of $A$?
(Second direction): Now suppose that $y\in \mathrm{Int}A$. If an open ball of $y$ is not in$ X-(X-A)^{-}$ then it must be in the closure of $X-A$
Question: is it correct to say that the open ball must overlap the set $X-A$? giving the contradiction $y$ is in both $A$ and $X-A$
With regard to the first direction. $X \backslash A \subseteq cl(X \backslash A) \Rightarrow X \backslash cl(X \backslash A) \subseteq X \backslash (X \backslash A) = A$.
B however is not necessarily a subset of A. For example: $X = [0,1]$ and $A = (0, \frac{1}{4}) \cup (\frac{3}{4}, 1) \Rightarrow B = (0, 1)$
With regard to the the second direction. $y \in int(A) \Rightarrow \exists r > 0 : B_r(y) \subset A \Rightarrow y \notin cl(X \backslash A).$
The proof follows from the theorem (easily proven with the de Morgans law) that $(cl(A))^c = int(A^c)$, where $\forall A \subset X \hspace{.3cm} A^c = X \backslash A$.
Thus we have $(cl(A^c))^c = int((A^c)^c) = int(A).$