Let $K$ be a field. I would like to prove that any algebraic extension $L$ of $K$ is separable iff ($\textrm{char} K = 0$ or $K = K^p$ in the case that $\textrm{char} K = p > 0$.
I have already proven the backward direction. In remains to show the forward direction. Thus, let $K$ be a field such that any algebraic extension $L$ of $K$ is separable. Let $\varphi : K \to K$ be the map $x \mapsto x^p$. This map can be shown to be injective. It remains to show that it's surjective. Let $\alpha \in K$. We want to show that $f(T) := T^p - \alpha$ has a root in $K$. Let $\beta \in \overline{K}$ be a root of $f$. Then $K \subseteq K(\beta)$ is an algebraic extension of $K$, and hence is separable. We can factor $f \in K[T]$ as the polynomial $f = (T - \beta)^p \in \overline{K}[T]$. I am not sure where to go from here. Could you guys help me finish?
Let me try the following:
Suppose $\text{char}(K)\neq 0$. Then $\text{char}(K)=q$ for some prime $q>0$.
Further suppose the Frobenius homomorphism $\sigma:x\to x^p$ is not bijection. Just as OP pointed out: the Frobenious map is always injective. Hence we must have $\sigma$ is not surjective. Aka, $\exists u \in K$ such that $u\notin \sigma(K)$. Define $f(x)=x^p-u \in K[x].$
Let $\alpha \in \bar{K}$ be a zero of $f(x)$, aka, $f(\alpha)=\alpha^p - u=0$. Hence via freshmen's dream: $$f(x)=x^p-u=x^p-\alpha^p=(x-\alpha)^p.$$ We claim that $f(x) \in K[x]$ irreducible over $K$. This holds because otherwise $\exists m \in (1,p]$ such that $(x-\alpha)^m = x^m-\alpha m x^{m-1}+... \in K[x]$ is an irreducible factor of $f(x)$. This further requires $\alpha m =\underbrace{\alpha + ... + \alpha}_{m\text{ items}}\in K$, which implies $m=p$.
Since $f(x)=x^p-u$ is irreducible over $K$, we know $\text{irr}(\alpha,K)=f(x).$ Note that by assumption, $K(\alpha)$ is a separable extension over $K$. Which requires $$1=\text{number of distinct zeros of }\text{irr}(\alpha,K)=\{K(\alpha): K\}=[K(\alpha):K]=\deg(\text{irr}(\alpha,K))=p.$$ Contradiction arises. Hence the Frobenius homomorphism must be a bijection at first place.