Munkres Topology, page 102, question 19:a 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:a] Show that $\text{Int } A$ and $\text{Bd } A$ are disjoint, and $\bar{A} = \text{Int } A \cup \text{Bd } A$.


MY ATTEMPT (via an EXAMPLE)
Suppose that I have the closed disk
$$A = \left\{(x,y) \in \mathbb{R}^2 | x^2 + y^2 \leq 1\right\}.$$
Clearly, since $A$ is closed, then I have $\bar{A} = A$.
I also have
$$\text{Int } A = \left\{(x,y) \in \mathbb{R}^2 | x^2 + y^2 < 1\right\},$$
and
$$\text{Bd } A = \left\{(x,y) \in \mathbb{R}^2 | x^2 + y^2 = 1\right\},$$
so that
$$\text{Int } A \cap \text{Bd } A = \emptyset$$
(i.e., $\text{Int } A$ and $\text{Bd } A$ are disjoint), and
$$\bar{A} = A = \text{Int } A \cup \text{Bd } A.$$
QUESTION

This gives some intuition for how to solve the problem in case $A$ is closed (although 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 for closed $A$?  How about the case when $A$ is open?

Added September 05 2017  Of course, a set can be both open and closed, or can be neither open nor closed!  Now I am more confused...  =(
 A: If $x\in\operatorname{Int}A$, then there is a neghborhood $N$ of $x$ such that $N\subset A$. Therefore, $N\cap(X\setminus A)=\emptyset$ and so $x\notin \overline{X\setminus A}$. This proves that $\operatorname{Int}A$ and $\operatorname{Bd}A$ are disjoint.
If $x\in\overline A$ and $x\notin\operatorname{Int}A$, then no neighborood of $x$ is a subset of $A$, which means that every neighborood of $x$ intersects $X\setminus A$. Therefore $x\in\operatorname{Bd}A$. So $\overline A\subset\operatorname{Int}A\cup\operatorname{Bd}A$.
Finally, if $x\in\operatorname{Int}A\cup\operatorname{Bd}A$, then


*

*if $x\in\operatorname{Int}A$, then $x\in A$ and therefore $x\in\overline A$;

*if $x\in\operatorname{Bd}A$, then, by definition, $x\in\overline A$.


Therefore, $\overline A=\operatorname{Int}A\cup\operatorname{Bd}A$.
A: The interior of $A$ is denoted $A^o$
We know that(or it can be proved if you want) that $$\overline{X-A}=X-A^o$$
Thus $$A^o \cap bd(A)=A^o \cap(\bar{A} \cap (X-A^o))= \emptyset$$
Now to prove the second statement use these: 

$x \in bd(A)$ iff for every   open neighbourhood $U$ of $x$ we have that $U \cap A \neq \emptyset$ and $U \cap (X-A) \neq \emptyset$

.

$x \in \bar{A}$ iff every open neighborhood of $x$ intersects $A$

