Can an open set in $\mathbb{R}$ be described as the disjoint union of uncountably many open intervals? The lecture notes I'm working through pose the question of whether any open set in $\mathbb{R}$ can be described as the disjoint union of uncountably many open intervals.
The problem doesn't state which topology on $\mathbb{R}$ I'm using. I know that any open set can be written as a union of open balls, and an open ball in $\mathbb{R}$ with its usual topology is an open interval, so I assume this is using the standard topology. (If it doesn't matter, I'd be interested in knowing.)
Intuitively, I believe the answer is no, but here's my attempt at a proof. It essentially mimics the argument that any collection of non-overlapping discs in $\mathbb{R}^2$ is countable.

Let $U \subset \mathbb{R}$ be open, and $\{G_{\alpha}\}_{\alpha \in I}$ an uncountable collection of disjoint, open intervals such that $U = \bigcup\limits_{\alpha \in I} G_{\alpha}$. As each $\mathbb{Q}$ is dense in $\mathbb{R}$, in each $G_{\alpha}$, we can choose a rational $q_{\alpha}$. As the $\{G_{\alpha}\}$ are disjoint, if $q_{i}$ lives in $G_i$ for some $i \in I$, then $q_i \not \in G_{\alpha}$ for every $\alpha \neq i$. Therefore, we can write an injection $f: I \to \mathbb{Q}$ sending $\alpha \mapsto q_{\alpha}$. As $f$ is injective, $I \cong f(I) \subset \mathbb{Q}$. As $\mathbb{Q}$ is countable and a subset of a countable set is countable, $f(I)$ is countable, so $I$ is countble, and we have a contradiction.

I'd appreciate some advice on how to improve the proof. I don't think I need contradiction, for example. I could just just take $U$ to be a union of disjoint open intervals -- which may not even exist; at least, I haven't proved that -- and then prove that $I$ must be countable. I also found myself having trouble using appropriate notation to distinguish between the variable $\alpha$ and the specific index $i$.
 A: If $\Bbb R$ has the standard order, let $G_i, i \in I$ be a pairwise disjoint family of open intervals , say $G_i = (a_i, b_i)$ with $a_i < b_i$ two reals for each $i \in I$.
For each $i \in I$ we can pick a rational $q_i \in \Bbb Q$ so that $a_i < q_i < b_i$. This has nothing to do with any topology but with the construction/definition of $\Bbb R$ from $\Bbb Q$.
It follows that $I \to q_i$ is an injective function from $I$ (by disjointness) into a countable set. Hence $I$ is at most countable.
There simply doesn't exist any uncountable disjoint family of open intervals in $\Bbb R$, answering the question independently of any topology you might put on $\Bbb R$...
A: As I stated in a comment you are basically proving that there cannot exist an uncountable family of disjoint open sets (you can easily extend your proof to arbitrary separable spaces). But we can in facht prove this without AOC:
Let F be a family of open disjoint sets.
We define for $q\in\mathbb Q$ the set $F_q = \{O\in F:q\in O\}$. The disjoint property then means that $F_q$ has at most one element. But for each $O\in F$ there exists at least on $q\in O$, so
$$ F = \bigcup_{q\in\mathbb Q} F_q$$
and is thus at most countable.
The trick here is that we do not in fact need any uniqueness of this $q$.
A: This is the direct proof you want:
Let $U$ be an open set in $\mathbb{R}$ and let us define an equivalence relation:
$$x\sim y\Leftrightarrow\exists I\subset U\text{ open interval such that }x,y\in I$$
This is easily proven to be an equivalence relation and thus we can consider $U/\sim$. In this space, the equivalence classes are given by the open intervals of $U$. If we denote as $I_x$ the open interval containing $x\in U$, let us prove that $|U/\sim|\leq\aleph_0$:
For every $x\in U$ there exists $a_x\in\mathbb{Q}\cap I_x$ (because $\mathbb{Q}$ is dense in $\mathbb{R}$) and $x\nsim y\Rightarrow I_x\cap I_y=\emptyset$. Thus, using the axiom of choice, we can define $\phi:\mathbb{R}/\sim\to\mathbb{Q}$ by $\phi(I_x)=a_x$. Trivially $x\nsim y\Rightarrow \phi(I_x)\neq \phi(I_y)$ and then $|U/\sim|\leq|\mathbb{Q}|\leq\aleph_0$, as we wanted to prove.
Although this is a direct proof the idea is more or less the same as what you did in your contradiction proof.
