Let me prove subset $A$ of $\mathbb{R}$ is open if and only if it is a countable union of open intervals.
For all $x \in A$ there is an $\epsilon > 0$ such that $(x-\epsilon,x+\epsilon)$ is an open interval contained in $A$. Now find rationals such that $r_{x} \in (x-\epsilon,x)$ and $s_{x} \in (x,x+\epsilon)$ and $A = \bigcup_{x \in A}(r_{x},s_{x})$. Note that the intervals with rational end points is less than $\mathbb{Q} \times \mathbb{Q}$. Obviously $\mathbb{Q} \times \mathbb{Q}$ is countable.
My doubt is that instead of selecting rationals can I select irrationals.