I'm struggling to understand the claim that if $E/F$ is a finite field extension, $F$ has no odd extensions, and $E$ has no extensions of degree $2$, then $F$ is perfect and $E$ is algebraically closed.
I was thinking, since $F$ has no nontrivial odd extensions, then every extension is divisible by $2$, and it's known from field theory that in this case every finite extension has degree a power of $2$.
In particular, every nontrivial finite extension of $E$ must be have degree a power of $2$. If any one of these extensions is Galois, then choosing a fixed field of a subgroup of a suitable index would give a field extension of $E$ of degree $2$, a contradiction. Thus $E$ would have no nontrivial finite extensions, implying it's algebraically closed. Is there a way to make this work that I'm missing?
I'm blank on how to see that $F$ is perfect? If $F$ is imperfect of characteristic $p$, then $F\neq F^p$. If $a\in F\setminus F^p$, then I recall that $X^{p^e}-a$ is irreducible, so adjoining any root would give an extension of $F$ of degree $p^e$. Since $F$ only has extensions of degree $F$, necessarily $\operatorname{char}(F)=0,2$. In the former case, $F$ is perfect, but if $\operatorname{char}(F)=2$?
Furthermore, if $\operatorname{char}(F)=0$, let $K/E$ be a finite extension. Since we're working over characteristic $0$, we can enlarge this to a Galois extension, denoted $\tilde{K}$, to get a tower $\tilde{K}/E/F$. So $\tilde{K}/E$ is Galois as well, but then as noted above, if this is not proper, $E$ will have a quadratic extension. Thus $\tilde{K}=E$, so $K=E$. Thus $E$ is algebraically closed.
So I feel the only remaining difficulty is when $\operatorname{char}(F)=2$.
Edit: Maybe it's a sledgehammer, but I think showing $E$ is algebraically closed without assuming characteristic $0$ is sufficient. For if $[E:F]=1$, then $E=F$ is perfect as all algebraically closed fields are. If not, since then $1<[E:F]<\infty$, Artin-Shreier implies $E=F(\sqrt{-1})$ and $\operatorname{char}(F)=0$, so $F$ is again perfect.