Separable closure is a field Let $K\subset L$ a field extension. Define
$K_s=\{x\in L : x \text{ is algebraic and separable over } K\}$
Show that $K_s$ is a subfield of $F$.
Attempt:
I tried to prove that $K(\alpha,\beta)/K$ is a separable extension when $\alpha,\beta$ are separable and algebraic (I know that $K(\alpha,\beta)/K$ is algebraic when $\alpha,\beta$ are algebraic). Then $\alpha\pm\beta, \alpha^{-1},\alpha\beta, \alpha/\beta$ are separable then $K_s$ is a field.
I have no idea how to prove that $K(\alpha)/K$ is separable if $\alpha$ is separable. Also, if I prove that, how to conclude? I tried to prove the following "If $K/F$  is separable and $L/K$  is separable then $L/F$  is separable" 
 A: 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.
A: 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.
A: I would like to add one important bit detail to Ewan Delanoy's answer : the one that for a tower of algebraic extensions $F \subset L \subset K$ and $N$ an algebraic closure of $K$, one has the equality $|\textbf{Hom}_F(K,N)| = |\textbf{Hom}_L(K,N)|\cdot |\textbf{Hom}_F(L,N)|$.
Here's a detailed proof :
We will describe a bijection between the set $|\textbf{Hom}_F(K,N)|$ and the product $|\textbf{Hom}_L(K,N)|\times |\textbf{Hom}_F(L,N)|$. To do so, for each element of $|\textbf{Hom}_F(K,N)|$, one easily obtains by restriction a unique element of $|\textbf{Hom}_F(L,N)|$ and an element of $|\textbf{Hom}_{L \to \sigma(L)}(K,N)|$, which is an element of $|\textbf{Hom}_F(K,N)|$ which sends $L \to \sigma(L)$.
Now define an equivalence relation on the elements of $|\textbf{Hom}_F(K,N)|$ by $\sigma \simeq \sigma' \iff \sigma|_L = \sigma'|_L$. For each of the classes, using the axiom of choice, choose an extension $\tilde{\sigma}$ of the mapping $\sigma(L) \to L$ to an isomorphism $N \to N$. For each $\sigma'$ in the class of $\sigma$, one obtains by post-composition by $\tilde{\sigma}$ a bijection $|\textbf{Hom}_{L \to \sigma(L)}(K,N)| \simeq |\textbf{Hom}_{L}(K,N)|$. We have thus described a mapping  $|\textbf{Hom}_F(K,N)| \to |\textbf{Hom}_L(K,N)|\times |\textbf{Hom}_F(L,N)|$. It is injective, since if $\sigma$ and $\sigma'$ are the same on $L$ then the second component is obtained by composition by the same isomorphism, and thus this forces $\sigma = \sigma'$. It is also surjective, since any element of the product $\sigma,\tau$ is obtained by taking the element of $\textbf{Hom}_F(K,N)$ defined by considering the class of $\tau$ in $\textbf{Hom}(K,N)$, and considering a pre-image of $\sigma$ through the bijection $|\textbf{Hom}_{L \to \tau(L)}(K,N)| \simeq |\textbf{Hom}_{L}(K,N)|$.
