I am trying to prove this statement in the characteristic $p>0$ case.
Every algebraic extension of a perfect field is separable and perfect.
This is stated as a corollary of Proposition V.6.11 in Lang.
Let $k$ be a field and let $K$ be a normal extension of $k$. Let $G$ be the group of automorphisms of $K$ that fix $k$. Let $K^G$ be the fixed field of $G$ acting on $K$, and $K_0$ the maximal separable subextension of $K$. Then $K^G$ is purely inseparable over $k$, $K$ is separable over $K^G$, $K_0\cap K^G =k$ and $K_0K^G=K$.
Lang doesn't write many details in the proof except that every finite extension is contained in a normal extension.
This is what I have so far. Suppose, $k=k^p$. Let $E$ be an algebraic extension of $k$. Let $\alpha \in E$. Now, $\alpha$ is contained in the field, say $K$ generated by the roots of the minimal polynomial of $\alpha$ over $k$ (and is hence normal as it's the splitting field of the minimal polynomial of $\alpha$). At this point, I don't know how to proceed. I am tempted to use the Frobenius map, but it does not fix every element of $k$ although it does stabilize it in this perfect case. Any hints will be appreciated. Thanks.