Show that the boundary of $A$ is empty iff $A$ is closed and open.

As you said $\bar{A} = A \cup \operatorname{Bd}(A)$, but there's a more stronger statement that you can and you should prove: $\bar{A} = \operatorname{Int}(A) \cup \operatorname{Bd}(A)$.

Now it should be fairly easy to prove that $\operatorname{Bd}(A) = \emptyset \iff \text{A is closed and open}$:

(1)If $\operatorname{Bd}(A)=\emptyset$

Then $\bar{A}=\operatorname{Int}(A)$ and since $\operatorname{Int}(A)\subset A \subset \bar{A}$ we conclude that $\operatorname{Int}(A) = A = \bar{A}$.

$\operatorname{Int}(A) = A$ shows that $A$ is open and $A = \bar{A}$ shows that $A$ is closed.

(2)If $A$ is closed and open.

A is open, then $A=\operatorname{Int}(A)$ and since $\bar{A} = \operatorname{Int}(A) \cup \operatorname{Bd}(A)$ we see that $\bar{A} = A \cup \operatorname{Bd}(A)$

But $A$ is also closed!, this means that $\bar{A} = A$ therefore $A = A \cup \operatorname{Bd}(A)$. Now remember that $\operatorname{Int}(A) \cap \operatorname{Bd}(A) = \emptyset$, then $A \cap \operatorname{Bd}(A) = \emptyset$.

Because of $A = A \cup \operatorname{Bd}(A)$ and $A \cap \operatorname{Bd}(A) = \emptyset$ we conclude $\operatorname{Bd}(A) = \emptyset$

$\emptyset=\partial A=\overline{A}\setminus \mathring{A}$ (the first is the hypothesis, the second a well-known property) together with $\mathring{A} \subset A\subset\overline{A} $ gives you $\mathring{A}=A=\overline{A}$ so that A is clopen (and from the first inequalities the other implication is obvious).

We know $\operatorname{Bdry}(A)=\operatorname{cl}(A)\cap \operatorname{cl}(X-A)$. So $\operatorname{Bdry}(A)=\operatorname{cl}(A)\cap (X-\operatorname{int}(A))$. Hence $\operatorname{Bdry}(A)=\operatorname{cl}(A)-\operatorname{int}(A)$. Now use this identity.