Let $K$ be a maximal subfield of $\mathbb C$ which does not contain $\sqrt{2}$. I've shown that $\mathbb C$ is an algebraic extension of $K$, and that the Galois group of any finite extension of $K$ is a cyclic $2$-group. I need to now prove that $[\mathbb C:K]$ is countable and not finite.
If $K$ contained the roots of unity, then the fact that the extension is infinite would follow from $x^{2^k}-2$ being irreducible over $K$ for all $k$, since if it were reducible, then some "proper" power of $\sqrt[2^k]{2}$ would be in $K$, and then it would follow that $\sqrt{2}\in K$, a contradiction. But of course, $K$ can't contain all roots of unity, since if it contained $i$ and $\sqrt{i}=\dfrac{1}{\sqrt{2}}(1+i)$, then $\sqrt{2}\in K$. So without the roots of unity, I'm no longer sure if $x^{2^k}-2$ is irreducible over $K$.
I guess I could try using the fact that any finite extension of $K$ is a cyclic $2$-extension to show that $\mathbb C$ is not a finite extension of $K$, but I'm not sure how to do that. Is it because $\mathbb C$ can't have a finite order automorphism of order greater than $2$?
As for the extension being countable, I want to say that $K\overline{\mathbb Q}=\mathbb C$ but I'm not sure of this. If this were not true, there would be a transcendental (over $\mathbb Q$) number $\alpha\notin K\overline{\mathbb Q}$. But $\sqrt{2}\in K(\alpha)$, and so $\alpha$ is a root of a polynomial in $K[x]$ (which is how I proved $\mathbb C$ is algebraic over $K$), but I'm not sure how to use this to show that $\alpha\in K\overline{\mathbb Q}$.