Is the absolute Galois Group of $\Bbb Q$ countable?

Let $I\subseteq \Bbb N$ be any subset and let $K_I=\Bbb Q(\{\sqrt{p_i}\}_{i\in I})$ where $p_i$ is the $i^{th}$ prime. Then there are precisely $2^{\Bbb N}$ in fact the Galois group of the compositum of this extension is exactly isomorphic to

$$\prod_{i\in\Bbb N}\Bbb Z/2\Bbb Z$$

and this of course indicates there are uncountably many elements in $\text{Gal}(\overline{\Bbb Q}/\Bbb Q)$

You can also do a simple cardinality argument using inverse limits, but that technology is a bit stronger.

Let me elaborate on Lubin's comment: there are no countably infinite Galois groups. This follows from a general statement about topological spaces: If $X$ is compact, Hausdorff with no isolated points, then $X$ is uncountable. For the proof, see here: https://proofwiki.org/wiki/Compact_Hausdorff_Space_with_no_Isolated_Points_is_Uncountable/Lemma