I am reading infinite Galois theory, and the motivation for introducing the Krull topology seems to be that this is the way in which we can solve the problem that the Galois correspondence fails. This is the counterexample given:
If $G=\text{Gal}( \bar{\mathbb{Q}}/\mathbb{Q})$ there are uncountably many subgroups of $G$ with index $2$, while the number of subfields of $\bar{\mathbb{Q}}$ of degree $2$ over $\mathbb{Q}$ is countable, thus there cannot be a bijection.
I understand why the number of subfields of $\bar{\mathbb{Q}}$ of degree $2$ over $\mathbb{Q}$ is countable (these are of the form $\mathbb{Q}(\alpha)$ with $\alpha$ a root of an irreducible polynomial of degree $2$, and there are countably many possible $\alpha$) but I don't get why there are uncountably many subgroups of $G$ with index $2$.
Any hints or help will be appreciated.
