Clarification requested about a claim in Abbott's "Understanding Analysis" In the opening sentences of chapter 3.5 on Baire's Theorem, Abbott writes:

The structure of open sets is relatively straightforward. Every open set is either a finite or countable union of open intervals.

I have looked backed through chapter 3 and I cannot find where he has proved this. Is it because the statement is completely trivial? If not, then it doesn't seem like Abbott's style to just assert something like this... I'm wondering if anyone could give me some clarification.
 A: If $A$ is an open set then every point, $\alpha$ is an interior point. So if for every $\alpha$ there is an open interval $I_\alpha$ so that $\alpha \in I_\alpha \subset A$.
So $\cup_{\alpha\in A}I_\alpha \subset A$. But also $A = \cup_{\alpha \in A} \subset \cup_{\alpha\in A}I_\alpha$ so $A =\cup_{\alpha\in A}I_\alpha \subset A$.
Now the rationals are dense so for each $I_\alpha$ there are rational numbers in it.  
So for each rational $q\in A$ there are (probably many) $I_\alpha$ so that $q\in I_\alpha$ but we consider the collection of collections of Intervals $U_q= \cup_{I_\alpha\ni q} I_\alpha$ And $A=\cup_{\alpha\in A} I_{\alpha} = \cup_{q\in A\cap \mathbb Q} U_q$.
Claim:  for each rational $q\in A$, $U_q$ is single interval.
Once shown, we are done.
If $x\in U_q, y\in U_q$ and $x< y$ then there is an $I_\alpha$ so that $x\in I_\alpha$ and as $I_\alpha$ is an interval all $w: \min(q,x)\le w \le \max(q,x)$ we will have $w\in I_\alpha$.  There is an $I_\beta$ so that $y\in I_\beta$.  So for all $w: \min(q,y) \le w\le \max(q,y)$ we wil have $w\in I_\beta$.  So for any $w: x \le w \le y$ then $\min(x,q)\le x \le w \le y\le \max(x,q)$ so $w\in U_q$.  So by definition $U_q$ is an Interval.
Oh,  I guess I have to show $U_q$ is open.
Also have to acknowledge I am allowing of unbounded intervals.  And as obviously there will be cases where $q\ne r$ but $U_q = U_r$ that $A= \cup_{a\in K}U_k$ for some $K \subset A\cap \mathbb Q$ and $K$ being "at most countable" is countable or finite.
...
Okay $U_q$ is open...  Let $x \in U_q$ the $x \in I_\alpha$ for some $\alpha \in A$ so ... yeah, that's obvious. $I_\alpha$ is open si there is in interval $I_x$ so that $x \in I_x \subset I_\alpha\subset U_q$ so $U_q$ is an open interval.
