Prove that Open Sets in $\mathbb{R}$ are The Disjoint Union of Open Intervals Without the Axioms of Choice

Question

There are several proofs I have seen of this, but they all seem to use choice subtely at some point. Is there any way to prove this without choice, or is it possibly unproveable?

Answer 1

The usual proof uses nothing of the axiom of choice.

Given a non-empty open set $U$, define an equivalence relation on $U$ by: $$x\sim y\iff\exists I\text{ an interval}, I\subseteq U, x,y\in I.$$

Now show that each equivalence class is open, simply by definition. Any point inside has an open interval around it contained in the equivalence class.

Therefore $U$ is the disjoint union of open intervals. And we can do more. We can use the fact that the rational numbers are countable to enumerate them, and choose a unique one from each interval, thus proving that the number of intervals is finite or $\aleph_0$ as well.