Excuse me can you see this question Every closed nowhere dense set is the frontier of an open set ...
I tried on it but i am not sure , I prove it as follows Let $A$ is closed nowhere dense set, take the open set $X\setminus A$ ..
Claim.. $\operatorname{Fr}(X\setminus A)=A$
$\operatorname{Fr}(X\setminus A) = \operatorname{Fr}(A)$ but $\operatorname{Fr}(A) \subseteq A$, since $A$ is closed so $\operatorname{Fr}(X\setminus A) \subseteq A$
To prove that $A \subseteq\operatorname{Fr}(X\setminus A)$ Let $x\in A$ s.t $A$ is closed nowhere dense , let $U(x,\epsilon)$ implies $U(x,\epsilon)\cap A \ne\varnothing$ .. but $U(x,\epsilon) =\varnothing$ since $A$ is nowhere dense Which is contradiction ($x \in U(x,\epsilon)$) so $U(x,\epsilon)\cap (X\setminus A)\ne\varnothing$ thus $x \in \operatorname{Fr}(X\setminus A)$ thus $A \subseteq \operatorname{Fr}(X\setminus A)$
The first part is okay. I think that you have the right idea for the second part, but what you’ve written is very hard to follow and partly wrong. In particular, there is no reason to think that $X$ is a metric space, so you can’t use $\epsilon$-balls. Fortunately, you don’t need them: open nbhds of any kind will do. Let $x\in A$, and let $U$ be an open nbhd of $x$. $A$ has empty interior, so $U\nsubseteq A$, and therefore $U\cap(X\setminus A)\ne\varnothing$. $U$ was an arbitrary open nbhd of $x$, so $x\in\operatorname{cl}(X\setminus A)$, and therefore $x\in\operatorname{Fr}(X\setminus A)$.