Find collection of countable open discs in $\mathbb{C}$ such that any open $U\subset \mathbb{C}$ is a union of such discs.

Question

I'm trying to find a collection S that is countable, and a collection of open discs $\{ D_i \}_{i\in S}$ such that any open $U\subset \mathbb{C}$ can be expressed as the union of such $D_i$'s. Now, given an open $U\in \mathbb{C}$, I can put $\mathbb{Q}^2 = S$ and then make a series of discs such that $\forall u\in U, u\in B_{\epsilon}(q),q\in \mathbb{Q}^2$ for some $\epsilon$, and $B_{\epsilon}(q)\subset U$. I call $D_q=B_{\epsilon}(q)$. That this is possible is easy to show as $\mathbb{Q}^2$ is dense in $\mathbb{C}$. This is the general idea for when I am given a $U$ and I want to find such a set $S$ and make these $D_i$. But how do I make a single collection of $\{D_i\}_{i\in S}$ such that ANY open $U\in \mathbb{C}$ can be written as the union of these $D_i$'s?

Edit: I actually seriously doubt that this is possible, so could the question I'm reading be wrong? Suppose this was possible, take $D_i$ for some $i$, but since it is a disk $D_i=B_{\epsilon}(x)$ for some $x$. But then consider $B_{\frac{\epsilon}{2}}(x)$

It's not entirely clear to me what you want to do. I gather that $S$ is a subset of $\mathbb C$? Then what does it mean that you use $i\in S$ as a subscript on $D_i$? That you want the open discs to be centred on the points $i\in S$, one disc per point? — joriki, Feb 05 '23 at 22:45
$S$ is simply an indexing set. I am making ${D_i}$ that is indexed by elements in S. For the rational radius, what if I have $D_r(q)$, and I simply put $U=D_{\frac{r}{2}}(q)$. Then this does not hold. — Goob, Feb 05 '23 at 22:50
I mean, let ${D_r(x+iy):x, y\in\mathbb{Q}, r\in\mathbb{Q}^+}$ be the collection. This is countable, and satisfies the condition. — Kangyeon Moon, Feb 05 '23 at 23:09
Does this answer your question? How to show that the set of open balls with rational centres and rational radii form a countable base for $\mathbb{R}^n$? — Anne Bauval, Feb 06 '23 at 00:03
Your title asks for a "collection of countable open disks such that blah" but that is not what you want. Indeed if it were, the answer would be that there is none, for no disk is countable. — Mariano Suárez-Álvarez, Feb 06 '23 at 05:22

Find collection of countable open discs in $\mathbb{C}$ such that any open $U\subset \mathbb{C}$ is a union of such discs.

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