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