Munkres Topology, page 102, question 19:b PROBLEM

If $A \subset X$, we define the boundary of $A$ by the equation
  $$\text{Bd } A = \bar{A} \cap \overline{X - A}.$$

[Munkres Topology, page 102, question 19:b] Show that
    $$\text{Bd } A = \emptyset \iff A \text{ is both open and closed}.$$


MY ATTEMPT (again by EXAMPLES)
The set $\mathbb{R}$ (with the usual topology) has empty boundary, and it is both open and closed.
The empty set also has empty boundary in this topology, and it is also both open and closed.
QUESTION

Again, this gives some intuition for how to solve the problem in special cases only. (Admittedly, I currently have trouble articulating a general proof [i.e., a proof that does not depend on specific examples, such as what I have given above]).  How do I solve the problem generally?

 A: By 19:a (Munkres Topology, page 102, question 19:a), we know
$$\overline{A} = \text{Bd} A \cup \text{Int} A.$$
Moreover we know that $\text{Int}A$ and $\text{Bd} A$ are disjoint. Further, $A$ is open iff $A = \text{Int} A$ and $A$ is closed iff $\overline A = A$.
So let's assume $A$ is open and closed, then $\overline A = \text{Int} A$ and since $\text{Int}A$ and $\text{Bd} A$ are disjoint, we obtain $\text{Bd} A = \emptyset$.
For the other direction, assume $\text{Bd} A = \emptyset$. Then we obtain $\overline A = \text{Int} A$. And since $\overline A \supseteq A \supseteq \text{Int}A$ we get $\overline A = \text{Int} A = A$ which shows that $A$ is open and closed.
A: Suppose that $bd(A)$ is empty. Then, we have to prove that $A$ is both closed and open. Instead, we will prove that both $A$ and $X-A$  are open.
Well, what does it mean for $A$ to be open? Given any element $x \in A$, there should be  an open set $U$ containing $x$, such that $U \subset A$. Suppose there is no such  $U$. This means, for every $U$ containing $x$, $U \cap (X-A) \neq \emptyset$. Note that this implies that $x \in \overline{X-A}$ (because every open neighbourhood of $x$ intersects $X-A$ non-trivially). But then, $x \in A \subset \overline A$, so $x$ must belong to $bd(A)$, which is a contradiction.
I urge you to show similarly that $X-A$ is open. It is very similar to the above argument.
Now, if $A$ is both open and closed, then any limit point  $x$ of $A$ lies within $A$. Furthermore, since $A$ is open, there is a neighbourhood $U$ of $x$ that is contained entirely in $A$, and hence disjoint from $A$, so that $x$ cannot possibly be a limit point of $X-A$. Repeat the same argument on the other side to show that the boundary of $A$ must be empty.  
