Solution: Let $A$ be both open and closed in $X$. Since $A$ is closed, $\bar{A} = A$ and as $A$ is open, the complement is closed and hence $\overline{X\setminus A} = X\setminus A$. Then the boundary of $A$ is given by \begin{align*} \partial A & = \overline{A} \cap \overline{X\setminus A} \\ & = A \cap (X\setminus A) \\ & = \emptyset. \end{align*}

On the other hand, let $\partial A = \emptyset $. Recall that if $x \in \partial A$, then every neighbourhood $U_x \subseteq X$ of $x$, $U_x \cap A \neq \emptyset $ and $U_x \cap (X\setminus A) \neq \emptyset $. Since $\partial A = \emptyset $, \begin{align*} & \forall\ x\in X, \quad \exists\ U_x \subseteq X \text{ such that } U_x \cap A = \emptyset \text{ or } U_x \cap (X\setminus A) = \emptyset \\ \implies & \forall\ x\in A, \quad \exists\ U_x \subseteq X \text{ such that } U_x \cap (X\setminus A) = \emptyset \\ \implies & A \text{ is open in } X. \end{align*} Similarly, we can show that $X\setminus A$ is open and hence $A$ is closed in $X$. Therefore, $A$ is both open and closed in $X$.