$\bar{\mathbb{F}}_p$ is not a finite degree extension of any proper subfield. Consider $\bar{\mathbb{F}}_p$, the algebraic closure of $\mathbb{F}_p$. I want to see that: for every proper subfield $K \leq \bar{\mathbb{F}_p}$, $\bar{\mathbb{F}}_p/K$ is not a finite extension.
It is known that, and can be somewhat easily shown that $\bar{\mathbb{F}}_p = \cup_{n \geq 1}\mathbb{F}_{p^n}$
Now, if any of the proper subfields have the form $\mathbb{F}_{p^n}$, it is easy enough to see that $\bar{\mathbb{F}}_p \neq \mathbb{F}_{p^n}(\alpha_1, \cdots, \alpha_m)$ for some $\alpha_i$, by going high up enough, i.e, to some big enough $m$ such that $\alpha_i \not \in \mathbb{F}_{p^m} \subseteq \bar{\mathbb{F}_p}$
The problem is characterizing the proper subfields. Is every subfield of $\bar{\mathbb{F}_p}$ going to have this form? Can we have an infinite intermediate subfield?

As Slade explained (+1) this follows either from the known structure of the automorphism group of $\overline{\Bbb{F}_p}$ or from a theorem of Artin & Schreier stating that any finite extension $\overline{K}/K$, with the bigger field algebraically closed, is of the form $\overline{K}=K(\sqrt{-1})$.
An elementary argument can also be given. Assume that $\overline{\Bbb{F}_p}/K$ is a finite extension. Then that extension is Galois. It is obviously normal, and it is also separable because every element of $\overline{\Bbb{F}_p}$ belongs to a finite field, and hence is a zero of a separable polynomial over the prime field (and hence also over $K$). By basic Galois theory this implies that $\overline{\Bbb{F}_p}$ has an automorphism of a finite order. But I shamelessly link to an old elementary answer of mine explaining that there are no such automorphisms.
