The crucial point is: It is not trivial at all to see that $\mathbb{Q}(\sqrt{p_1},\dotsc,\sqrt{p_n})$ has degree exactly $2^n$ (only $\leq 2^n$ is clear) over $\mathbb{Q}$, or equivalently, that $p_i \notin \mathbb{Q}(\sqrt{p_1},\dotsc,\sqrt{p_{i-1}})$ for all $i \leq n$.
Here is what you can do: Let $m_1,\dotsc,m_r$ be coprime and square free numbers $>1$ (for example distinct primes). Then I claim that $K_r := \mathbb{Q}(\sqrt{m_1},\dotsc,\sqrt{m_r})$ has degree $2^r$ over $\mathbb{Q}$.
It is enough to prove $\sqrt{m_{i+1}} \notin K_i$ for all $i < r$. But it turns out one should show something stronger: For $i<r$ and $i < j_1 < \dotsc < j_s \leq r$ we have $\sqrt{m_{j_1} \cdot \dotsc \cdot m_{j_s}} \notin K_i$. Now this can be shown by induction on $i$; I leave it for you as an exercise.
Now the rest is easy: $K_r$ is the splitting field of $(x^2-m_1) \cdot \cdots \cdot (x^2-m_r)$, hence normal, and everything in characteristic $0$ is separable. Thus $K_r$ is Galois over $\mathbb{Q}$; let $G$ be the Galois group. Since the conjugates of $\sqrt{m_i}$ are $\pm \sqrt{m_i}$, there is a homomorphism $\alpha : G \to \{\pm 1\}^r$ given by $\alpha(g) \sqrt{m_i} = g(\sqrt{m_i})$. Clearly it is injective. Since $G$ has order $2^r$, it is also surjective. Thus we have an isomorphism $G \cong \{\pm 1\}^r$. Explicitly, the Galois automorphisms are exactly as you've described them, they are given by $\sqrt{m_i} \mapsto \varepsilon_i \sqrt{m_i}$, where $\varepsilon_i = \pm 1$.
Remark: There are shorter proofs, using Kummer Theory.
Here is a more general situation. Let $F$ be a field, $a \in F^\times$, and assume $X^n - a$ is irreducible over $F$.
(1) We want to show for each $d|n$ that $X^d - a$ is irreducible over $F$.
(2) Writing $\sqrt[n]{a}$ as notation for a root of $X^n - a$, assume any $n$th roots of unity in $F(\sqrt[n]{a})$ in fact lie in $F$.
(Example: $F = {\mathbf Q}$ and $a > 0$, so ${\mathbf Q}(\sqrt[n]{a})$ is isomorphic to a subfield of ${\mathbf R}$, which makes it clear that the only roots of unity at all in ${\mathbf Q}(\sqrt[n]{a})$ are $\pm 1$, which lie in ${\mathbf Q}$.) We want to show, for each $d|n$, that the only field between $F$ and $F(\sqrt[n]{a})$ of degree $d$ is $F(\sqrt[d]{a})$, where $\sqrt[d]{a} := \sqrt[n]{a}^{n/d}$.
Proof of (1): Write $\sqrt[d]{a}$ for $\sqrt[n]{a}^{n/d}$, so $\sqrt[d]{a}$ is a root of $X^d - a$. In the tower $F \subset F(\sqrt[d]{a}) \subset F(\sqrt[n]{a})$, we have $[F(\sqrt[d]{a}):F] \leq d$ and $[F(\sqrt[n]{a}):F(\sqrt[d]{a})] \leq n/d$, since $\sqrt[d]{a}$ is a root of $X^d - a \in F[X]$ and $\sqrt[n]{a}$ is a root of $X^{n/d} - \sqrt[d]{a} \in F(\sqrt[d]{a})[X]$. Because $$[F(\sqrt[n]{a}):F] = [F(\sqrt[n]{a}):F(\sqrt[d]{a})][F(\sqrt[d]{a}):F]$$
and we assume the left side is $n$, it follows that our upper bounds for the terms on the right must be equalities. In particular, $[F(\sqrt[d]{a}):F] = d$, so $X^d - a$ must be irreducible over $F$ (it has a root with degree $d$ over $F$).
Proof of (2): Let $d|n$ and suppose $E$ is a field with $F \subset E \subset F(\sqrt[n]{a})$ and $[E:F] = d$. To prove $E = F(\sqrt[d]{a})$, it suffices to show $\sqrt[d]{a} \in E$, since that would give us $F(\sqrt[d]{a}) \subset E$ and we already saw in (1) that $F(\sqrt[d]{a})$ has degree $d$ over $F$, so the containment $F(\sqrt[d]{a}) \subset E$ would have to be an equality.
Let $f(X)$ be the minimal polynomial of $\sqrt[n]{a}$ over $E$, so $f(X)|(X^n - a)$ and $\deg f = n/d$. Any two roots of $f(X)$ are $n$th roots of $a$, and thus have a ratio that is an $n$th root of unity, so in terms of the one root $\sqrt[n]{a}$ we can write any other root of $f(X)$ as $\zeta\sqrt[n]{a}$ for some $n$th root of unity $\zeta$. (I am not making any assumptions about the $n$th roots of unity being distinct, in case $F$ has positive characteristic, and these individual $n$th roots of unity need not lie in $F(\sqrt[n]{a})$.) In a splitting field, the factorization of $f(X)$ is $\prod_{i \in I} (X - \zeta_i\sqrt[n]{a})$ for some $n$th roots of unity $\zeta_i$ ($I$ is just an index set). The constant term of $f(X)$ is in $E$, so $(\prod_{i \in I} \zeta_i)\sqrt[n]{a}^{n/d} \in E$. Therefore $(\prod_{i \in I} \zeta_i)\sqrt[n]{a}^{n/d} \in F(\sqrt[n]{a})$, so $\prod_{i \in I} \zeta_i \in F(\sqrt[n]{a})$. The only $n$th roots of unity in $F(\sqrt[n]{a})$ are, by hypothesis, in $F$, so $\prod_{i \in I} \zeta_i \in F \subset E$. Therefore $\sqrt[n]{a}^{n/d} = \sqrt[d]{a}$ is in $E$, so we're done.
To see an example of this not involving the rational numbers, let $k$ be a field and $F = k(t)$, the rational functions over $k$ in one indeterminate. The polynomial $X^n - t$ is irreducible over $k(t)$, since it is Eisenstein at $t$. We let $\sqrt[n]{t}$ denote one root of $X^n - t$, so $F(\sqrt[n]{t}) = k(\sqrt[n]{t})$ has degree $n$ over $k(t)$. All roots of unity in $k(\sqrt[n]{t})$ -- not just $n$th roots of unity -- are in $k$, because $k(\sqrt[n]{t})$ is a rational function field in one indeterminate over $k$ (it's $k$-isomorphic to $k(t)$) and there's a general theorem that any roots of unity in a rational function field over $k$ are in the constant field $k$. By the above work, the only fields between $k(t)$ and $k(\sqrt[n]{t})$ are $k(\sqrt[d]{t})$ for $d|n$.
An example where the hypothesis that all $n$th roots of unity in $F(\sqrt[n]{a})$ are in $F$ is false, and the conclusion is nevertheless true, is $F = {\mathbf Q}(i)$, $a = 2$, and $n = 8$: $[{\mathbf Q}(i,\sqrt[8]{2}):{\mathbf Q}(i)] = 8$. The extension ${\mathbf Q}(i,\sqrt[8]{2})/{\mathbf Q}(i)$ is Galois with a cyclic Galois group, but not all 8th roots of unity are in ${\mathbf Q}(i)$.
Best Answer
Let $L = \mathbb Q(\sqrt{-1},\sqrt{2},\sqrt{3},\ldots)$ and let $G=\textrm{Gal}(L/{\mathbb Q})$.
Facts.
Let me briefly explain the 2nd, 3rd, and 6th of these.
The set of subfields of $L$ that are finite extensions of $\mathbb Q$ has cardinality $\aleph_0$.
Let $\mathcal F$ be the set of intermediate extensions $\mathbb Q\leq F\leq L$ of finite degree over $\mathbb Q$. Map $L$ to $\mathcal F$ by $\alpha\mapsto \mathbb Q[\alpha]$. This is a surjective map from a countable set $L$ onto $\mathcal F$, so $\mathcal F$ is countable. Since $\mathcal F$ contains infinitely many distinct members, e.g. $\mathbb Q[\sqrt{-1}], \mathbb Q[\sqrt{2}], \mathbb Q[\sqrt{3}], \ldots$, we must have $|\mathcal F|= \aleph_0$.
The set of all subfields of $L$ has cardinality $2^{\aleph_0}$.
Since $|L|=\aleph_0$, the set $L$ has $2^{\aleph_0}$ subsets. The number of subfields must be $\leq 2^{\aleph_0}$. However, the argument in the second paragraph of the question statement shows that the number of subfields is at least $2^{\aleph_0}$, so we have equality.
The set of index-$2$ subgroups of $G$ has cardinality $2^{2^{\aleph_0}}$.
$G\cong \mathbb Z_2^{\aleph_0}$ (see Example 3.10 of Conrad's notes). This means $G$ can be viewed as an $\mathbb F_2$-vector space of cardinality $2^{\aleph_0}$. When the cardinality of a vector space is infinite and strictly exceeds the cardinality of the field, then the cardinality equals the dimension, so $G$ must have an $\mathbb F_2$-basis $\mathcal B$ of size $2^{\aleph_0}$. Different surjective functions $f\colon {\mathcal B}\to \mathbb F_2$ extend to different surjective homomorphisms $\overline{f}\colon G\to \mathbb F_2$, and these different surjective homomorphisms necessarily have different kernels. Each kernel is a subgroup of $G$ of index $2$. Since there are $2^{2^{\aleph_0}}$-many surjective functions $f\colon {\mathcal B}\to \mathbb F_2$, there are at least this many subgroups of $G$ of index $2$. There can't be more since $G$ only has $2^{2^{\aleph_0}}$-many subsets.
Now to get to the question: is there a mistake in Example 3.10 of Conrad's notes?
I found no mathematical mistake in Example 3.10. I did not read the rest of the notes. I found it slightly misleading that Conrad would write [[$\textrm{Gal}(L/{\mathbb Q})$ has uncountably many subgroups of order 2. At the same time, $L$ has only countably many subfields of each $2$-power degree over $\mathbb Q$.]] Writing this way could lead the reader to believe that that there is a reason to compare subgroups of finite order in $\textrm{Gal}(L/{\mathbb Q})$ to intermediate extensions of $L/\mathbb Q$ of finite degree over $\mathbb Q$. Instead one should compare subgroups of finite index in $\textrm{Gal}(L/{\mathbb Q})$ to extensions of finite degree over $\mathbb Q$. I think it would have been clearer to write [[$\textrm{Gal}(L/{\mathbb Q})$ has $2^{2^{\aleph_0}}$-many subgroups of index $2$. At the same time, $L$ has only $\aleph_0$-many subfields of degree $2$ over $\mathbb Q$.]]
Next, Conrad writes in the concluding two lines of Example 3.10 [[$L$ has only countably many subfields of each $2$-power degree over $\mathbb Q$. Therefore the subfields of $L$ and the subgroups of $\textrm{Gal}(L/{\mathbb Q})$ do not have the same cardinality.]] The word Therefore connects mathematical claims. These claims are correct. But the final claim is not a consequence of former claims, so the word Therefore is not the best choice. In fact, Conrad does not determine the number of subfields of $L$ nor the number of subgroups of $\textrm{Gal}(L/{\mathbb Q})$ in this example, so there is no place for Therefore in what he has written.