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.

In a locally connected space $X$, all connected components of open sets are open. This is in fact equivalent to being locally connected.
Proof: (one direction) let $O$ be an open subset of a locally connected space $X$. Let $C$ be a component of $O$ (as a (sub)space in its own right). Let $x \in C$. Then let $U_x$ be a connected neighbourhood of $x$ in $X$ such that $U_x \subset O$, which can be done as $O$ is open and the connected neighbourhoods form a local base. Then $U_x,C \subset O$ are both connected and intersect (in $x$) so their union $U_x \cup C \subset O$ is a connected subset of $O$ containing $x$, so by maximality of components $U_x \cup C \subset C$. But then $U_x$ witnesses that $x$ is an interior point of $C$, and this shows all points of $C$ are interior points, hence $C$ is open (in either $X$ or $O$, that's equivalent).
Now $\mathbb{R}$ is locally connected (open intervals form a local base of connected sets) and so every open set if a disjoint union of its components, which are open connected subsets of $\mathbb{R}$, hence are open intervals (potentially of infinite "length", i.e. segments). That there are countably many of them at most, follows from the already given "rational in every interval" argument. 
