If we let $K$ be a maximal subfield of $\mathbb C$ that doesn't contain $\sqrt{2}$, then if $L$ is any finite extension of $K$, then its Galois with a cyclic 2-group. I want to use this to conclude that $[\mathbb C:K]$ is countable and not finite. I know that $\mathbb C$ is algebraic over $K$.
I'm not quite sure how to do this. I sort of have an idea for the not finite part: if it were finite, then $\mathbb C$ would be a cyclic extension of $K$, and so it would have a unique index 2 subfield. If I can prove that the only index $2$ subfield of $\mathbb C$ is $\mathbb R$, then I think I can finish the argument, since $K$ "should" contain imaginary elements (though I'm not too sure how to prove this either).
For countable, does the fact that every complex number is a root of a polynomial in $K[x]$ and the fact that $[K[x]:K]$ is countable work?
See Algebraically closed fields with proper maximal subfields at MathOverflow for a succinct answer (shorter than the one I wrote, until I recognized it as one I had read over there, looked up the reference, and deleted my answer).