Prove that if $[F(\alpha):F]$ is odd then $F(\alpha)=F(\alpha^2)$
Your proof is correct but you don't need contradiction. If $F(\alpha) \neq F(\alpha^2)$ then $\alpha \notin F(\alpha^2)$. Thus consider the tower $[F(\alpha) : F(\alpha^2)][F(\alpha^2) : F]$. The first factor is at most two (because $\alpha$ is a root of $x^2 - \alpha^2$ and it is at least two because $\alpha \notin F(\alpha^2)$. By the tower law it follows $[F(\alpha) : F]$ is even.