TITLE: Every open set in the real line is the countable union of disjoint intervals
QUESTION [6 upvotes]: Prove that every open set in $\mathbb{R}^1$ is the union of an at most countable collection of disjoint intervals.

Proof:
Let $\mathbb{R}^1\supset G$ is an open set. Then $\forall$ $x\in G$ $\exists \varepsilon_x>0:$ $I_x=(x-\varepsilon_x, x+\varepsilon_x)\subset G$. Then $$G=\cup_{x\in G}I_x.$$
But $\mathbb{R}^1$ is separable then it has a countable base then any open cover has a countable subcover. Then $$G=\cup_{i\geqslant}I_i.$$
How to turn these intervals to disjoint intervals?

REPLY [2 votes]: For the record, an alternative topological approach that may be useful to someone:
Take your open set $U$. Consider its partition in its connected components. Since they are connected, they must be intervals.
Now, it is easy to see that a disjoint family of intervals must be enumerable: take any rational in its midsts as the enumeration.