Let $(X,d)$ be a metric space and let $A\subset X$. For $x \in X$, define $$d(x,A)=\text{inf}\; \{d(x,y) \;\vert\; y \in A \}$$
If $$\partial A=\{x \in X \;\vert \; d(x,A)=0\} \cap \{x \in X \;\vert \; d(x,X\setminus A)=0\} $$ then how to prove $\partial A$ is closed for any $A \subset X$ ?
My try: Since $f:x \mapsto d(x,A)$ is continuous and $g:x \mapsto d(x,X \setminus A)$ is continuous, so $$\partial A=f^{-1}\Bigl(\{0\}\Bigr) \cap g^{-1}\Bigl(\{0\}\Bigr)$$ and intersection of two closed set is closed, $\partial A$ is closed!
Is my proof right? If no, what is the mistake and how to approach this ?
The proof is fine if you already know that $f$ is continuous. But in any space (not just metric) we have that $$\partial A = \overline{A} \cap \overline{X\setminus A}$$ which is an intersection of closed sets and thus closed. The $f$ is somewhat overkill.