Let $A$ be closed. In general, the boundary of a set is closed. So it suffices to show that $\partial A$ has empty interior.

Let $U\subset\partial A$ be open. Since $\partial A\subset \overline{A}=A$ this implies that $U\subset A$ Since $\operatorname{int}(A)$ is the largest open subset of $A$ we must have $U\subset\operatorname{int}(A)$ Therefore $U\subset \partial A \cap \operatorname{int}(A)$ But $\partial A \cap \operatorname{int}(A)=(\overline{A}-\operatorname{int}(A))\cap\operatorname{int}(A)=\emptyset$ so $U=\emptyset$