Classification of an open set in real Prove that open set in real line can be represented as ar most countable disjoint union of open intervals. 
I know that this question repeated many times in MSE but let me ask the following question. I found in internet the following nice proof but one moment  seems to me weird:

Let $U \subseteq R$ be open and let $x \in U$. Either $x$ is rational or irrational. If $x$ is rational, define
  \begin{align}I_x = \bigcup\limits_{\substack{I\text{ an open interval} \\ x~\in~I~\subseteq~U}} I,\end{align}
  which, as a union of non-disjoint open intervals (each $I$ contains $x$),  is an open interval subset to $U$. If $x$ is irrational, by openness of $U$ there is $\varepsilon > 0$ such that $(x - \varepsilon, x + \varepsilon) \subseteq U$, and there exists rational $y \in (x - \varepsilon, x + \varepsilon) \subseteq I_y$ (by the definition of $I_y$). Hence $x \in I_y$. So any $x \in U$ is in $I_q$ for some $q \in U \cap \mathbb{Q}$, and so
  \begin{align}U \subseteq \bigcup\limits_{q~\in~U \cap~\mathbb{Q}} I_q.\end{align}
  But $I_q \subseteq U$ for each $q \in U \cap \mathbb{Q}$; thus
  \begin{align}U = \bigcup\limits_{q~\in~U \cap~\mathbb{Q}} I_q, \end{align}
  which is a countable union of open intervals.

My question is the following: How to show that these $I_q$ are disjoint. Suppose that $x\in I_p\cap I_q$? How to show that $I_p\cup I_q\subset I_p$ and $\subset I_q$?
 A: Here’s a fairly detailed argument.
Suppose that $x,y\in U\cap\Bbb Q$, and $I_x\cap I_y\ne\varnothing$. Since $I_x\cap I_y\ne\varnothing$, there is a $z\in I_x\cap I_y$. By the definition of $I_x$ there is an open interval $J_0$ such that $x\in J_0\subseteq U$ and $z\in J_0$. Similarly, by the definition of $I_y$ there is an open interval $J_1$ such that $y\in J_1\subseteq I$ and $z\in J_1$. Let $J_2=J_0\cup J_1$; since the open intervals $J_0$ and $J_1$ have the point $z$ in common, $J_2$ is an open interval, and $x,y\in J_2\subseteq U$.
Now let $u\in I_x$ be arbitrary; there just be an open interval $J_3$ such that $x\in J_3\subseteq U$ and $u\in J_3$. Let $J=J_2\cup J_3$; the open intervals $J_2$ and $J_3$ have the point $x$ in common, so $J$ is an open interval, and clearly $y\in J\subseteq U$, so $J\subseteq I_y$. Thus, $u\in J\subseteq I_y$, and since $u$ was an arbitrary point of $I_x$, we’ve shown that $I_x\subseteq I_y$. 
Exactly the same argument with the rôles of $x$ and $y$ reversed shows that $I_y\subseteq I_x$, so $I_x=I_y$. In other words, for any $x,y\in U\cap\Bbb Q$, either $I_x$ and $I_y$ are disjoint, or they are identical.
