proving every nonempty open set is the disjoint union of a countable collection of open intervals i'm studying real analysis with royden, and i've looked up similar qeustions and answers but i couldn't get the exact answer that i need. 
i don't need the whole process of proof , and i confused with certain phrase. 
First let me show you what i read. 
royden
i know the procedure in this proof, but i don't know why {I$_x$} is countable. 
they explain this in line 12 , but i can't understand it. 
especially, i think, when i try to construct {I$_x$} , there exist repeated open intervals for certain interval in open subset $O$ , then the number of open interval of {I$_x$} must be finite. 
then how can we say that {I$_x$} is countable?
with specific example , i understood I$_x$ for [3,10] as a interval of open set $O$ then, I$_7$ and I$_8$ is same , so we write only one of them into {I$_x$} , it means unless $O$ has countable interval, {I$_x$} cannot be countable ..
what do i missed? or misunderstood? let me know 

Each $I_x$ contains a rational number say $q_x$.
Since the $I_x$ are disjoint, each $q_x$ belongs to exactly one $I_x$.
This produces a bijection between the sets $\{I_x\}$ and $\{q_x\}$, which is a subset of the rationals and hence countable.
