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?

I miswrote what I had intended, and will write a corrected version here.  Someday I will try to post an updated copy of the file where you found it. My aim there was to suggest a mismatch between subgroups and intermediate fields without aiming to make such a statement absolutely precise since, as you noted, the Krull topology was not yet introduced.
Let $L = \mathbf Q(\sqrt{-1},\sqrt{2},\sqrt{3},\sqrt{5},\sqrt{7},\dots)$, so
$G := {\rm Gal}(L/\mathbf Q) = \prod \{\pm 1\}$, which is a countable direct product of copies of
the group $\{\pm 1\}$ and $G$ is abelian.
Every quadratic extension of $\mathbf Q$ is in $L$, and there are countably many such extensions.
We expect by Galois theory that these extensions are in bijection with the subgroups of $G$ with index 2, but
we'll show there are uncountably many index-2 subgroups of $G$.
To each subgroup $H \subset G$ with index 2, $G/H \cong \mathbf F_2$ as groups in exactly one way, so
reduction mod $H$ gives us a nonzero group homomorphism $G \to \mathbf F_2$ with kernel $H$. The number of these $H$ is the number of nonzero group homomorphisms $G \to \mathbf F_2$.
Since all elements of $G$ square to the identity, we can think of $G$ as an $\mathbf F_2$-vector space
and think of nonzero homomorphisms $G \to \mathbf F_2$ as nonzero elements of the dual space
${\rm Hom}_{\mathbf F_2}(G,\mathbf F_2)$.  The group $G$ is uncountable (that is explained in other answers here), and this implies
the $\mathbf F_2$-dual space of $G$ is uncountable (also explained elsewhere on this page).
Therefore the number of index-2 subgroups of $G$ is uncountable.
