Exercise 19, Section 17 of Munkres’ Topology

Solution 1:

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..