Every open set in $\mathbb{R}$ is the disjoint union of open intervals I know this is a standard question and that I can easily find solutions on this site or elsewhere. However, I came up with a proposed proof and would like someone to review it for me. If this is known, my apologies.
Let $A_{\alpha}$ be a family of open intervals. Given $\alpha, \alpha'$ we say $A_{\alpha} \sim A_{\alpha'}$ if there exist $\alpha_{1}, ..., \alpha_{n}$ such that $A_{\alpha} \cap A_{\alpha_{1}} \neq \emptyset, ..., A_{\alpha_{n}} \cap A_{\alpha'} \neq \emptyset$.
We see that $\sim$ is an equivalence relation. Consider $A$ to be an equivalence class and $F$ to be the union of all elements of $A$. Considering $a = \inf F$, $b = \sup F$ (where $a, b$ take values in the extended reals), we claim $F = (a, b)$.
Let $a < x < b$. It suffices to see $x \in F$. This is clear since there exists $\alpha, \alpha'$ with $A_{\alpha}, A_{\alpha'}$ in $A$ such that $A_{\alpha}$ contains points smaller than $x$ (since $x$ is not the infimum) and $A_{\alpha'}$ contains points greater than $x$. Taking $\alpha_{1}, ..., \alpha_{n}$ as in the definition we see that for some $i$, $A_{\alpha_{i}}$ contains $x$.
If it is not true that $A_{\alpha} \sim A_{\alpha'}$ then $A_{\alpha} \cap A_{\alpha'} = \emptyset$. Thus each $F$ is disjoint, and the union of $A_{\alpha}$ is the union of the $F$. Therefore any open subset of $\mathbb{R}$ is the union of disjoint open intervals
 A: The idea is sound, but the implementation could be better. Here’s a fairly careful write-up missing only a few details.
Let $U$ be a non-empty open subset of $\Bbb R$, and let $\mathscr{I}$ be the family of all open intervals contained in $U$. For $I,J\in\mathscr{I}$ write $I\sim J$ 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 $k=0,\dots,n-1$; clearly $\sim$ is an equivalence relation on $\mathscr{I}$. For $I\in\mathscr{I}$ let $[I]$ be the $\sim$-equivalence class of $I$, and let $U_I=\bigcup[I]$; clearly $U_I$ is open.
Suppose that $U_I\cap U_J\ne\varnothing$ for some $I,J\in\mathscr{I}$; then there are $I'\in[I]$ and $J'\in[J]$ such that $I'\cap J'\ne\varnothing$. Clearly $I\sim I'\sim J'\sim J$, so $I\sim J$, and $[I]=[J]$. Thus, $\{U_I:I\in\mathscr{I}\}$ is a partition of $U$ into open subsets.
Suppose that $I,J\in\mathscr{I}$ and $I\cap U\ne\varnothing$; then $I\cup J\in\mathscr{I}$. (Why?) An easy induction on $n$ then shows that if $I_0,\dots,I_n\in\mathscr{I}$, and $I_k\cap I_{k+1}\ne\varnothing$ for $k=0,\dots,n-1$, then $\bigcup_{k=0}^nI_k\in\mathscr{I}$. Fix $I\in\mathscr{I}$, and suppose that $x,y\in U_I$ with $x<y$. Then there are $I_x,I_y\in[I]$ such that $x\in I_x$ and $y\in I_y$. Now $I_x\sim I\sim I_y$, so there are $I_x=I_0,\dots,I_n=I_y\in[I]$ such that $I_k\cap I_{k+1}\ne\varnothing$ for $k=0,\dots,n-1$. Let $J=\bigcup_{k=0}^nI_k$; then $x,y\in J\in[I]$, so $[x,y]\subseteq J\subseteq U_I$. This shows that $U_I$ is order-convex: if $x,y\in U_I$ and $x<z<y$, then $z\in U_I$. 
The order-convex subsets of $\Bbb R$ are precisely the intervals, so the open, order-convex subsets of $\Bbb R$ are the open intervals (including the open rays), so $\{U_I:I\in\mathscr{I}\}$ is a partition of $U$ into open intervals, as desired.

You can also prove it by defining an equivalence relation directly on $U$: for $x,y\in U$, $x\sim y$ iff $\big[\min\{x,y\},\max\{x,y\}\big]\subseteq U$. This makes some of the argument a little easier and some a little harder; overall it’s probably pretty much a wash.
A: Your fourth paragraph is a little bit sketchy with $A_\alpha \sim A_{\alpha'}$ via the $\alpha_i$ implying some $\alpha_i$ contains $x$. The gaps would be filled in with an induction starting from the union of two overlapping intervals being an interval, so that you can read $A_\alpha\sim A_{\alpha'}$ as 
$B=\left[A_\alpha\cup \bigcup_{i=1}^{n-1} A_{alpha_i}\right]\sim A_{\alpha'}$ where $B$ and $A_{\alpha'}$ are overlapping intervals. So there's no real problem there.
I thought your last paragraph was wrong, but I guess you were thinking without specifying that $\cup_\alpha A_\alpha=U$ represents an arbitrary open set. It does, but you should start the proof with $U$, not with $A_\alpha$.
