Elements of $K$ which are separable over $F$ form a subfield of $K$? I'm trying to prove the following statement:

If $K$ is an extension of $F$ prove that the set of elements in $K$ which are separable over $F$ forms a subfield of $K$.

I have a proof for the set of algebraic elements in $K$ forming a subfield, but I'm stuck with the separable elements.
I'm assuming I should be splitting this into cases of characteristic = 0 and characteristic $\neq$ 0. Any help? I'm at a loss.
 A: Characteristic $0$ is immediate, since all elements are separable (irreducible polynomial has no repeated roots).
For characteristic $p$, let $\alpha$ and $\beta$ be separable. The characteristic polynomial of $\beta$ over $F(\alpha)$ divides the characteristic polynomial of $\beta$ over $F$, hence has no repeated roots; so $\beta$ is separable over $F(\alpha)$. Now consider the tower $F\subseteq F(\alpha+\beta)\subseteq F(\alpha,\beta)$. Looking at the separable degrees, we have:
$$[F(\alpha,\beta):F]_s = [F(\alpha,\beta),F(\alpha+\beta)]_s[F(\alpha+\beta):F]_s.$$
But 
$$[F(\alpha,\beta):F]_s = [F(\alpha,\beta):F(\alpha)]_s[F(\alpha):F]_s = [F(\alpha,\beta):F(\alpha)][F(\alpha):F] = [F(\alpha,\beta):F],$$ 
and $[F(\alpha,\beta):F(\alpha+\beta)]_s\leq [F(\alpha,\beta):F(\alpha+\beta)]$, $[F(\alpha+\beta):F]_s \leq [F(\alpha+\beta):F]$. So in order to get
$$\begin{align*}
[F(\alpha,\beta):F]&=[F(\alpha,\beta):F]_s \\
&= [F(\alpha,\beta),F(\alpha+\beta)]_s[F(\alpha+\beta):F]_s\\
&\leq [F(\alpha,\beta):F(\alpha+\beta)][F(\alpha+\beta):F]\\ 
&= [F(\alpha,\beta):F],\end{align*}$$
we must have equality throughout, hence $\alpha+\beta$ is separable over $F$. The same argument holds for $\alpha\beta$.
