I have a question concerning the characterization of open sets in a metric space.
Per definition a set O in a metric space (X,d) is open, if it is a neighborhood of each of its points. In the book "Introduction to Topology" by Mendelson (https://archive.org/details/IntroductionToTopology/page/n33/mode/2up) there is the following characterization:
Theorem (5.2): A subset O of a metric space (X,d) is an open set if and only if it is the union of open balls.
The proof of one direction is as follows: Suppose O is open. Then for each $a \in O$, there is an open ball $B(a;\delta_a) \subset O$. Therefore $O = \bigcup_{a \in O} B(a;\delta_a)$ is a union of open balls.
Now I wonder: Does this argument require the axiom of choice?
You need a 'choice function' which assigns to each $a \in O$, an element $\delta_a$ from the (non empty) set of all $\delta$'s where $B(a;\delta) \subset O$.
No, it doesn't require the axiom of choice. You can, for each $a\in O$, take the smallest $n\in\mathbb N$ such that $B\left(a;\frac1n\right)\subset O$.