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. 

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Let $U\subseteq\mathbb R$ open. It is enough to write $U$ as a disjoint union of open intervals.
For each $x\in U$, we define $\alpha_x=\inf\{\alpha\in\mathbb R:(\alpha,x+\epsilon)\subseteq U, \text{ for some }\epsilon>0\}$ and $\beta_x=\sup\{\beta\in\mathbb R:(\alpha_x,\beta)\subseteq U\}$.
Then $\displaystyle U=\bigcup_{x\in U}(\alpha_x,\beta_x)$ where $\{(\alpha_x,\beta_x):x\in U\}$ is a disjoint family of open intervals.
The intervals appearing in the union are disjoint in the sense that every time $x,y\in U$ with $x<y$, then either $(\alpha_x,\beta_x)=(\alpha_y,\beta_y)$ holds, or $(\alpha_x,\beta_x)\cap(\alpha_y,\beta_y)$ is empty. To see this, suppose $(\alpha_x,\beta_x)\cap(\alpha_y,\beta_y)$ has an element. We claim that $[x,y]\subseteq U$. (For if $x<t<y$ with $t\not\in U$, then $\beta_x\leq t\leq \alpha_y$.)
But if $[x,y]\subseteq U$, then both $\alpha_x<x$ and $\alpha_y<x$, and both $y<\beta_x$ and $y<\beta_y$. Hence   $\alpha_x$ and $\alpha_y$ can be expressed as $$\alpha_x=\inf\{\alpha\leq x:(\alpha,x+\epsilon)\subseteq U, \text{ for some }\epsilon>0\},$$ $$\alpha_y=\inf\{\overline\alpha\leq x:(\overline\alpha,y+\overline\epsilon)\subseteq U, \text{ for some }\overline\epsilon>0\},$$ and these are the same; so then also $\beta_x$ and $\beta_y$ are the same.
