Consider F―p, the algebraic closure of Fp. I want to see that: for every proper...

Step 1
As Slade explained ${\displaystyle (+1)}$ this follows either from the known structure of the automorphism group of ${\displaystyle \overline{F_p}}$ or from a theorem of Artin & Schreier stating that any finite extension ${\displaystyle \frac{\bar{K}}{K}}$, with the bigger field algebraically closed, is of the form ${\displaystyle \bar{K}=K\left(\sqrt{-1}\right)}$.
Step 2
An elementary argument can also be given. Assume that ${\displaystyle \frac{\overline{F_p}}{K}}$ is a finite extension. Then that extension is Galois. It is obviously normal, and it is also separable because every element of ${\displaystyle \overline{F_p}}$ belongs to a finite field, and hence is a zero of a separable polynomial over the ' field (and hence also over K). By basic Galois theory this implies that ${\displaystyle \overline{F_p}}$ has an automorphism of a finite order.

The Galois group here is isomorphic to ${\displaystyle G=\hat{Z}}$, 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.