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 $\alpha\beta$.