Any open subset of $\Bbb R$ is a countable union of disjoint open intervals Let $U$ be an open set in $\mathbb R$. Then $U$ is a countable union of disjoint intervals. 


This question has probably been asked. However, I am not interested in just getting the answer to it. Rather, I am interested in collecting as many different proofs of it which are as diverse as possible. A professor told me that there are many. So, I invite everyone who has seen proofs of this fact to share them with the community. I think it is a result worth knowing how to prove in many different ways and having a post that combines as many of them as possible will, no doubt, be quite useful. After two days, I will place a bounty on this question to attract as many people as possible. Of course, any comments, corrections, suggestions, links to papers/notes etc. are more than welcome.
Here’s one to get things started.

Let $U$ be a non-empty open subset of $\Bbb R$. For $x,y\in U$ define $x\sim y$ iff $\big[\min\{x,y\},\max\{x,y\}\big]\subseteq U$. It’s easily checked that $\sim$ is an equivalence relation on $U$ whose equivalence classes are pairwise disjoint open intervals in $\Bbb R$. (The term interval here includes unbounded intervals, i.e., rays.) Let $\mathscr{I}$ be the set of $\sim$-classes. Clearly $U=\bigcup_{I \in \mathscr{I}} I$. For each $I\in\mathscr{I}$ choose a rational $q_I\in I$; the map $\mathscr{I}\to\Bbb Q:I\mapsto q_I$ is injective, so $\mathscr{I}$ is countable.

A variant of the same basic idea is to let $\mathscr{I}$ be the set of open intervals that are subsets of $U$. For $I,J\in\mathscr{I}$ define $I\sim J$ iff there are $I_0=I,I_1,\dots,I_n=J\in\mathscr{I}$ such that $I_k\cap I_{k+1}\ne\varnothing$ for $k=0,\dots,n-1$. Then $\sim$ is an equivalence relation on $\mathscr{I}$. For $I\in\mathscr{I}$ let $[I]$ be the $\sim$-class of $I$. Then $\left\{\bigcup[I]:I\in\mathscr{I}\right\}$ is a decomposition of $U$ into pairwise disjoint open intervals.

Both of these arguments generalize to any LOTS (= Linearly Ordered Topological Space), i.e., any linearly ordered set $\langle X,\le\rangle$ with the topology generated by the subbase of open rays $(\leftarrow,x)$ and $(x,\to)$: if $U$ is a non-empty open subset of $X$, then $U$ is the union of a family of pairwise disjoint open intervals. In general the family need not be countable, of course.
The following is certainly not the quickest approach to a proof, but when this question was first posed to me in class, my first intuition was to use some elementary graph theory:



Let $U$ be an open set of $\mathbb{R}$. As we know, $\mathbb{R}$ has a countable basis $\mathcal{B}$ comprised of connected open sets and so we may write $U=\bigcup_{n\in I} U_n$, where for each $n$ we have $U_n\in\mathcal{B}$ and $I$ is some countable index set. 

Let $G$ be the intersection graph of $\{U_n\}$. That is to say, the vertex set of $G$ is simply $\{U_n\}$ and there is an edge between $U_i$ and $U_j$ iff they have nonempty intersection. It's easy to convince yourself that:


This graph must have countably many graphically-connected components (otherwise we'd have uncountably many vertices which is impossible).
The intersection graph of $A\subseteq\{U_n\}$ is graphically-connected iff for any two $V,W\in A$ there is a sequence $V=U_{n_1},U_{n_2},\ldots,U_{n_k}=W$ such that $U_{n_i}\cap U_{n_{i+1}}\neq\varnothing$. 
The union $\bigcup A$ is a connected set of $\mathbb{R}$ whenever the intersection graph of $A$ is graphically-connected. 


Thus, when we take the union of all the vertices within a graphically-connected component, for every component, we obtain countably-many connected open sets. The union of these sets is of course $U$ itself. Since the connected open sets of $\mathbb{R}$ are intervals (including rays), we're done.



Side Note: This would also work in $\mathbb{R}^n$ or in general, any topological space $X$ that has a countable basis comprised of connected sets. Well, so long as we replace countable union of disjoint open intervals with countable union of disjoint open connected sets.