Is every rigid field perfect? A field is rigid iff its automorphism group is trivial.
A field $F$ is perfect iff all irreducibles in $F[x]$ are separable.

Is every rigid field perfect?
 A: The answer is no.
The natural place to look for a rigid field is in the function field of a generic, high genus smooth curve. For instance, over the complex numbers, Hurwiz's automorphism theorem guarantees that a compact Riemann surface of genus $g>1$ has at most $84(g-1)$ automorphisms. Having at least one nontrivial automorphism, for a Riemann surface of genus $g>2$, should be viewed as a very special property (genus $2$ is not sufficient, because genus $2$ curves are all hyperelliptic, and therefore admit an involution).
This paper of Bjorn Poonen shows that for any field $k$ and any integer $g \geq 3$, there exists a smooth curve $X$ of genus $g$ over $k$ such that $X$ has no nontrivial automorphisms over $\overline{k}$ (so, a fortiori it has no automorphisms over $k$). 
Taking $k=\mathbf F_p$, it follows that the function field $\mathbf F_p(X)$ of such a curve $X/\mathbf F_p$ is rigid. However, being a finite extension of the rational function field $\mathbf F_p(T)$, it cannot be perfect, or else it would contain all $p$-power roots of $T$, and would have infinite degree over $\mathbf F_p(T)$.
