From (1), every open subset of $\Bbb R$ is at most a countable union of open intervals. The converse is also true: any countable union of open intervals is an open set. However, I want every subset of $\Bbb R$. What if we allow countable sets in our countable union?
For example, any closed interval $[a,b]$ can be written as the union:
$$ [a, b] = \{ a \} \cup (a, b) \cup \{ b \} $$
For this reason, I think very many subsets of $\Bbb R$ can be described this way. On my other hands, I don't believe the set of irrational numbers can be described this way, since it cannot contain any open sets (any open interval contains rationals) not can it be a union of countable sets (it is uncountable).
Edit: I asked only about this set's cardinality, but I am really interested in more specific details and how it compares to the entire power set of $\Bbb R$. I know this set has at least the cardinality of the continuum since it contains as a proper subset the collection of all open sets, which also has that cardinality.
Best Answer
The empty set is countable. Therefore, if $x$ is any real number then $I_x:=(x, x+1)$ is a union of an open interval and a countable set. For distinct real numbers $x, y$, $I_x, I_y$ are distinct. So the cardinality of the set that you are looking at has cardinality at least that of $\mathbb{R}$. It cannot have cardinality greater than that of $\mathbb{R}$ because, as you mentioned, open sets are countable unions of open intervals.
You may find this link helpful https://math.dartmouth.edu/archive/m103f08/public_html/borel-sets-soln.pdf