Exercise 19, Section 17 of Munkres’ Topology 
If $A\subseteq X$, we define the boundary of $A$ by the equation $\partial A=\overline{A}\cap \overline{X-A}$. (a)Show that $A^{\circ}$ and $\partial A$ are disjoint, and $\overline{A}= A^{\circ} \cup \partial A$. (b) Show that $\partial A=\phi \Leftrightarrow A$ is both open and closed. (c) Show that $U$ is open $\Leftrightarrow$ $\partial U= \overline{U}- U$.

I’m assuming, we know $\overline{X-A}= X-A^{\circ}$ fact. If you want, you can prove it.
My attempt: (a) $A^{\circ} \cap \partial A= A^{\circ} \cap (\overline{A} \cap \overline{X-A})=(\overline{A} \cap A^{\circ})\cap (X-A^{\circ})$, by associative law and above fact. Since $A^{\circ} \subseteq \overline{A}$, we have $\overline{A} \cap A^{\circ}= A^{\circ}$. Thus, $A^{\circ} \cap \partial A=\phi$.
$A^{\circ} \cup \partial A= A^{\circ} \cup (\overline{A} \cap \overline{X-A})= (A^{\circ} \cup \overline{A}) \cap (A^{\circ} \cup \overline{X-A})= \overline{A} \cap X= \overline{A}$, by De Morgan's law and above fact.
(b) Suppose $A$ is both open and closed. So $\overline{A}=A$ and $\overline{X-A}=X-A$. Thus $\partial A= A\cap (X-A)=\phi$.
Now suppose $\overline{A}\cap \overline{X-A}=\phi$. If $A$ is open, then $X-A$ is closed. So $\overline{X-A}= X-A$. Since $\partial A$ is empty, we have $\overline{A}\subseteq X-(X-A)=A$. Thus $\overline{A}=A$, $A$ is closed. Similarly result holds, if $A$ is closed. Is this proof correct?
(c) first suppose $U$ is open. Then $\partial U= \overline{U}\cap \overline{X-U}= \overline{U}\cap (X-U$), since $U$ is open. By Ex 2(g) of section 1, $(\overline{U} \cap X) - (\overline{U} \cap U) = \overline{U} - U$. Our desired result.
Conversely $\overline{U}-U = \partial U= \overline{U} \cap \overline{X-U}= \overline{U}\cap (X-U^{\circ})=\overline{U}-(\overline{U}\cap U^\circ)$. So $\overline{U}\cap U^\circ =U=U^\circ$. Thus $U$ is open. Is this proof correct?
 A: By definition, if $x \in \partial A = \overline{A} \cap \overline{X-A}$ has the property that every neighbourhood of $x$ intersects $A$ and $X-A$ as well. So it's clear that such an $x$ can never be in $A^\circ$ for then it has a neighbourhood that stays inside $A$ and so misses $X-A$ entirely.
This implies $\partial A \cap A^\circ = \emptyset$.
If $A$ is clopen, a point $x \in X$ is in $A$ and thus in $A^\circ$ and so not in $\partial A$ or it is in $X-A = (X-A)^\circ$ which is also disjoint from $\partial A$ (as $\partial A = \partial (X-A)$ from the definition and $(a)$ applies twice) so no point of $X$ can be in $\partial A$ and $\partial A = \emptyset$.
If $\partial A = \emptyset$ this means that if $x \in A$, $x \notin \partial A$ so $x \notin \overline{X-A}$, so for some neighbourhood $U$ of $x$, $U \cap (X-A)=\emptyset$ and so $U \subseteq A$ and $x \in A^\circ$ and so $A$ is open. The symmetric reasoning can be held to show $X-A$ is open so $A$ is closed and so $A$ is clopen.
Your reasoning for this implication is incorrect: if $A$ is open it is closed when $\partial A=\emptyset$. But you have to show for sure that $A$ is clopen, and it's not true that a set is always open or closed...
For (c), if you know that $\overline{X-A}=X-A^\circ$ we already have that $\partial U = \overline{U}-U^\circ$ for any $U$, which equals $\overline{U}-U$ when $U$ is open. If $\partial U = \overline{U}-U$ also equals $\overline{U}-U^{\circ} $ it follows that $U=U^\circ$ and so $U$ is open. Your computations seem overly complicated. Simplify..
A: All the questions can be answered easily using $\partial(A) =cl(A) \setminus int(A) $.
A point is a boundary point if it is neither in interior not in exterior.
Since you have already proved $\partial A=\overline{A}\cap \overline{X-A}$
and
$\overline{X-A}=X\setminus A^{\circ}$
Then it follows immediately that
$\partial A=\overline{A}\setminus A^{\circ}$
Now $(a), (b), (c) $ all three follows immediately.
