Separable closure is a field

So you’re asking for a crash course on separable extensions. It would go something like this : Let us work in an algebraic closure $C$ of $K$.

Lemma 1 Let $P\in K[X]$ be an irreducible polynomial. Then $P$ is not separable iff $P'=0$.

Proof of lemma 1 Suppose $P$ is inseparable, then there is $\alpha \in C$ such that $\alpha$ is a multiple root of $P$, so $(X-\alpha)^2$ divides $P$. Writing $P=(X-\alpha)^2Q$ and differentiating, we see that $X-\alpha$ divides $P'$. If we call $G$ the gcd of $P$ and $P'$, then $G$ annihilates $\alpha$. Since $P$ is irreducible, we must have $G=0$ and this is only possible if $P'=0$.

Conversely, suppose that $P'=0$. Let $\alpha$ be any root of $P$. We can write $P=(X-\alpha)Q$, and hence $P'=Q+(X-\alpha)Q'$. We deduce $Q(\alpha)=0$, so $\alpha$ is amultiple root of $P$, so $P$ is not separable. QED

Corollary of lemma 1 In characteristic zero, every irreducible polynomial is separable.

We henceforth assume that we are in characteristic $p > 0$. We denote ${\sf Hom}_K(L,C)$ the set of all field homomorphisms $L \to C$ that fix pointwise every element of $K$.

Lemma 2 One has $|{\sf Hom}_K(K(\alpha),C)| \leq [K(\alpha):K]$, with equality iff $\alpha$ is separable over $K$.

Proof of lemma 2 Any $\sigma \in {\sf Hom}_K(K(\alpha),C)$ is uniquely defined by its value at $\alpha$, and this value must be a conjugate of $\alpha$, which yields the claimed inequality. Conversely, for each conjugate $\bar{\alpha}$ of $\alpha$, there is a unique $\sigma \in {\sf Hom}_K(K(\alpha),C)$ sending $\alpha$ to $\bar{\alpha}$, QED.

Corollary 1 of lemma 2 If we have a tower of extensions $L_1/L_2/\ldots /L_r$ and each $L_i$ is $L_{i-1}(\alpha_i)$ for some $\alpha_i$, then $|{\sf Hom}_{L_r}(L_1,C)| \leq [L_1:L_r]$, with equality iff each $\alpha_i$ is separable over $L_{i-1}$.

Corollary 2 (of Corollary 1 above) If we have an extension $L/K$ of finite degree, then $|{\sf Hom}_K(L,C)| \leq [L:K]$, with equality iff $L$ is separable over $K$.

All the properties you need follow from this last corollary.

As Ewan Delanoy, first note the following result

Lemma Let F be a field and $f(x)\in F[x]$ be an irreducible polynomial with $deg f(x)\geqslant 1$.Then (1) if $char(F)=0$, then $f(x)$ has no multiple roots; (2) if $char(F)=p>0$, then $f(x)$ has multiple roots iff $f(x)=g(x^p)$ for some $g(x)\in F[x]$.

Then the follwing steps will answer your question:

Step 1 Any algebraic extension of an characteristic zero field is separable.

By above lemma.

Step 2 Let $E/F$ be a field extension with $char(F)=p>0$, then an element $\alpha\in E$ is separable over $F$ iff $F(\alpha)=F(\alpha^p)$.

By disproof and above lemma

Step 3 If $\alpha$ is separable over $F$, $\beta$ is separable over $F(\alpha)$, then $\beta$ is separable over $F$.

Assume that $char(F)=p>0$, then show that $F(\beta)=F(\beta^p)$.

Step 4 Let $E/F$ be an algebraic extension and $S\subseteq E$ be a subset, then $F(S)/F$ is separable iff every element of $S$ is separable.

By inductive step to the case of Step 3.