Topology: Reducing Open Set as Union of Open Balls to Countable Union

To prove that any open subset $A$ of $\Bbb{R}^2$ is a countable union of open disks, you need to take the union over all points $x \in A \cap \Bbb{Q}^2$ of all open disks $B(x, \delta)$ such that $\delta \in \Bbb{Q}$ and $B(x, \delta) \subseteq A$. If you just picked one $\delta$ for each $x$, then you might indeed omit some points of $A$.

For example,, take $A = \Bbb{R}^2$, and, for each $x \in A \cap \Bbb{Q}^2 = \Bbb{Q}^2$, pick a rational number $\delta_x > 0$ such that $d(x, (\sqrt{2}, 0)) < \delta_x$, then $(\sqrt{2}, 0)$ is not in any of the disks $B(x, \delta_x)$.