Show that any open cover of a subset of the plane has a countable subcover.

Question

This is my idea.

In $R^2$ $Q$ is dense in $R$ So pick rational numbers in the subset and their neighborhood can cover the subset. $Q$ is countable so the subcovers are countable. And in $C$, we can correspond $a+bi$ to $b/a$

But I think that is not enough to prove. How to solve this problem?

Answer 1

Let's talk just about open covers of $\Bbb R$ for simplicity. Yes, $\Bbb Q$ is countable, so if we have a family of open sets that covers $\Bbb R$ then some countable subfamily covers $\Bbb Q$. But that doesn't do what's required here, because an open cover of $\Bbb Q$ need not cover $\Bbb R$.

Reasonable thing to try but it doesn't work. Hint for something that does work: $\Bbb R$ is a countable union of compact sets...