Proposition 5.3 ("Real Analysis and Foundations", 2nd ed., by Steven Krantz)
Supposed that we have a set $U$=($-A$,$A$), where $A$ is a positive real number. If we partition this set into union of disjoint open sets, then it will create a natural singleton as the boundary point of adjacent open sets.
Take for example, we define an equiv. relation where a~b iff sign(a)=sign(b). Then we can partition $U$=($-A$,$0$) $\cup$ {$0$} $\cup$ ($0$,$A$). But then {$0$} is the closed singleton set mentioned above.
So how true again is this proposition or there is a warning underlying it that this is not supposed to be done?
Edit: Some parts that are originally confused were deleted.
To address your title, No. Just because a set can be decomposed into countable pairwise disjoint sets does not mean it can be decomposed into countable pairwise disjoint open/closed intervals.
The clearest example of this is a Vitali set. I will not define the set here, however, it is an uncountably infinite set $V$ such that for any two elements $x,y \in V$,
$$x-y \not\in \mathbb{Q}$$
If this set had a decomposition into countable pairwise open/closed intervals, one of the intervals must be uncountably infinite. But any uncountably infinite interval will contain 2 rational numbers. Their difference will be a rational number, and so this interval is not part of our decomposition (contradiction).