Does $\mathbb{Q}(\sqrt{-1}, \sqrt{2},\sqrt{3},\sqrt{5},\sqrt{7},\ldots)$ have countably many subfields? According to Example 3.10 of these notes, the field $L = \mathbb{Q}(\sqrt{-1}, \sqrt{2},\sqrt{3},\sqrt{5},\sqrt{7},\ldots)$, where we adjoin $\sqrt{p}$ for every prime $p$ (and $p=-1$) has only countably many subfields. I think that this is only true if we require the subfields to have finite degree over $\mathbb{Q}$.
If we do not specify finite degree, then for any element $(a_{-1}, a_2,a_3,a_5,\ldots) \in \prod_p \{0,1\}$, we get a subfield $E = \mathbb{Q}(a_{-1}\sqrt{-1}, a_2\sqrt{2}, a_3\sqrt{3},\ldots)$. This defines an injection from $\prod_p \{0,1\}$ to $\operatorname{Gal}(L/\mathbb{Q})$, so the latter is uncountable.
Given that Keith Conrad and I disagree, the conditional probability that I am wrong is high. Can anyone spot any errors in my reasoning?

There is no error, but you misread the exercice. The finite degree subfields don't correspond to the finite index subgroups but to the open finite index subgroups. The point of this exercice is this subtlety which is essential when dealing with infinite Galois extensions.
