In a metric space $X$, the boundary of an open set is the set of all limit points of a discrete set.
Actually, it is in the exercise of Willard, and it states that it requires the axiom of choice.
I thought an example, $S= \bigcup_{k \in \Bbb N} \left(\frac{1}{2k}, \frac{1}{2k-1}\right)$ is an open set in $\Bbb R$ and $\operatorname{Bd}(S)= \left\{ \frac1k \mid k \in \Bbb N \right\} \cup \{0\}$, and there is a discrete set, namely $T=\left\{\frac1m+\frac1n \mid m,n \in \Bbb N \right\}$ with derived set of $T$ being $\operatorname{Bd}(S)$.
And, for any open set $U$, $\operatorname{Bd}(U)= \operatorname{cl}(U)-U$.
But how to show existence of such a set like $T$ using the known facts. And, it can be guessed that the Choice Axiom may be used in showing existence of $T$ , but how?
This note on Topology Atlas, by Abhijit Dasgupta, gives a complete proof. The proof uses AC in the form of Zorn's lemma. The simpler proof sketch at the end, using paracompactness of a metric space, uses AC very blatantly, and IIRC already the fact that a metric space is paracompact requires a form of choice, in that it can fail in models without AC. I'm not sure if the original statement can fail in models without choice. (My hunch: it probably can).