$\bar{\mathbb{F}}_p$ is not a finite degree extension of any proper subfield.

Solution 1:

The Galois group here is isomorphic to $G=\hat{\mathbb{Z}}$, the profinite integers. Since $G$ is torsion-free, there are no finite index subfields.

This also follow from Artin-Schreier, though I don't know offhand of an extremely clean and short proof in the case of finite fields.

Solution 2:

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.