An open subset of $\mathbb{R}$ is an at-most-countable union of disjoint open intervals

Question

We have the classic result that an open subset of $\mathbb{R}$ is an at-most-countable union of disjoint open intervals.

In trying to prove this, I found the following line of reasoning, and I would like to know if its correct, because it's quite different from the other proofs I have read.

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Let S be an open subset of R.

Thus for every $a$ in $S$, $a$ is contained in some open ball $B$ which is contained in $S$.

In particular $a$ is contained in an open interval which is contained in $S$.

For each $a$, let's associate the maximal such interval $I_a=(\alpha, \beta)$ where $\alpha$ and $\beta$ live in $\mathbb{R} \bigcup \{\pm\infty\}$

$\alpha=\inf\{y: (y,a] \subset S\}$

$\beta=\sup\{y: [a,y) \subset S\}$

Now we have that $S$ is the union of $I_a$ over all $a$ in S. Furthermore if $I_{a_1}$ and $I_{a_2}$ are distinct, then they must be disjoint (otherwise we would contradict the maximality of at least one of them).

Hence $S$ is a disjoint union of open intervals.

Finally, each such interval contains a rational number which is not contained in any other such interval; if there were uncountably many intervals in the union, there would be uncountably many rationals, a contradiction.

Thus $S$ is an at-most-countable union of disjoint open intervals.

Answer 1

Your proof works. Perhaps the only thing which needs a bit more elaboration is why such "maximal" intervals exist. The only proper way to think about maximality is with respect to set inclusion, in other words for $a \in S$, $I_a$ is the unique open interval which contains every open interval contained in $S$ and containing $a$. So we define $I_a$ as follows: $$I_a = \bigcup_{I \text{ is an open sub-interval of $S$ and $a \in I$}} I$$

We know that the union of arbitrarily many connected sets (i.e. intervals) which have at least one point in common is also connected, so $I_a$ is an interval. It is open as the union of open sets.