I came up with the following proof that open sets can be written as a countable union of (not necessarily disjoint) open intervals, but I’m uncertain about one step I took. I’ve looked at Any open subset of $\Bbb R$ is a at most countable union of disjoint open intervals. [Collecting Proofs], but I not sure if my proof is among them.
Recall that the definition of open subset is that for every point $x$ in an open subset $U$, there exists some $\delta_x$ such that the neighborhood around the point, $(x-\delta_x, x+\delta_x)$ lies fully within $U$. That means $U$ can be written as $$U=\bigcup_{x\in U}(x-\delta_x, x+\delta_x)$$ which you can verify yourself by proving both $\subseteq$ and $\supseteq$. This does not guarantee a countable union, so we are not done. Let's make the following tweak: find a number, $\delta_x'\leq \delta_x$ such that $x-\delta_x'$ is rational. If $x-\delta_x' = x_1-\delta_{x_1}'$ for some $x_1\neq x$, then replace both intervals with the interval $(x-\delta_x', \max\{x+\delta_x', x_1+\delta_{x_1}'\})$. If we union over all these intervals, we get countably many open intervals (because the rationals are countable) that union to $U$: $$U= \bigcup_{x\in U}(x-\delta_x',\max\{\ldots\})$$ and we are done.
My concern is about the step where I combine potentially uncountably many intervals into one using $\max$. Is this allowed? Is the overall proof valid? If not, is there an easy fix, or should I scrap the proof?