How can I explain to my professor his argument invokes the AC? This is not the standard definition, but my topology professor restricted contexts in metric spaces.

Definition
An open set $U$ in a metric space $X$ is a subset of $X$ such that the interior of it and $U$ Itself are identical. (Interior point $x$ of a subset $A$ is a point of which there exists a ball containing $x$, contained in $A$.

With this definition, He showed that "Every open set in a metric space is a union of balls."
Below is his argument:

Assume that $U$ is open.
Let $x\in U$.
Then, by definition, there exists $r(x)$ such that $B(x,r(x))\subset U$, so that $U=\bigcup_{x\in U} B(x,r(x))$.

I'm not asking that this can be resolved easily without Axiom of Choice, but I'm asking how do I explain this to my professor his argument invokes Axiom of Choice. I told him that choosing $r(x)$ for each $x$ is a consequence of the axiom of choice, but he does not seem to understand me. Please help.
 A: My suggestion is not to bother your professor with this very much.
There are theorems whose choiceless proofs are very different than their choice-using proofs. There are theorems whose usual proof can be easily modified by showing that the arbitrary choice has some easy canonical choice instead.
When it is the former case, I agree that it might be worth insisting that a particular proof is using, or not using, the axiom of choice. But in the latter case it usually amount to unnecessary nitpicking (and that can always get people to dislike you).
In principle, you are right. When we say "some $r(x)$" rather than "The least $n$ such that ..." then we secretly invoke the axiom of choice. However since we can prove that the axiom of choice for this family of sets is true, there's little harm in this. (Not to mention that a lot of the time people use the axiom of choice in so many other places without noticing, that this one place where it's not a big deal shouldn't be different too.)
A: Setting aside the question of whether his argument uses the axiom of choice, the result doesn’t depend on it, and the proof is basically the same. Define $r(x)=\frac{1}{n}$, where $n$ is the smallest positive integer for which $B(x,\frac{1}{n})\subseteq U$.
