I am reading this PDF and concerned about Lemma 3. I will copy it here for completeness.
Lemma 3. Assume that $K \subseteq E \subseteq L$ is a tower of finite field extensions. If an element $\alpha$ of $L$ is separable over $K$, it is separable over $E$.
Proof. Let $f(x)$ be the minimal polynomial of $\alpha$ over $K$. Then of course $f(x) \in E[x]$, but it is not necessarily irreducible there. We may therefore factor $f(x) = g(x)h(x)$ over $E$, where $g(x)$ is the minimal polynomial of $\alpha$ over $E$ and where $h(\alpha) \neq 0$. Applying the product rule, we find $$f'(x) = g'(x)h(x) + g(x)h'(x)$$ Hence if $f'(\alpha) = 0$, then $g'(\alpha) = 0$ (since $h(\alpha) \neq 0$ and $g(\alpha) = 0$).
How do we know that $h(\alpha) = 0$? It must be that there is some property of $f$ and $g$ which forces them to have $\alpha$ as a root with equal multiplicity.
I thought for a second that it was perhaps because every minimal polynomial of $\alpha$ has $\alpha$ as a root with multiplicity $1$, but this is not so. Consider any nonperfect field $F$ of characteristic $p > 0$ and take some $\alpha \in F$ that does not have a $p$th root. Then the polynomial
$$f(x) = x^p - \alpha$$
has exactly one root with multiplicity $p$ over its splitting field. So, what else could it be? I've never seen this claim used before and, to be honest, the proof given above seems suspiciously simple.
Also, if this property does hold, I'd appreciate if anyone could given a proof which avoids Galois theory, as I have not yet gotten to that point of study.