Separability over intersection of intermediate fields

Let $$E\mathop{/}F$$ be a finite-degree normal extension of fields. Let $$K$$ and $$L$$ be intermediate fields (that is, $$F\subseteq K\subseteq E$$ and $$F\subseteq L\subseteq E$$) such that $$E$$ is separable over both $$K$$ and $$L$$. Show that $$E$$ is separable over $$K\cap L$$.

So I was thinking that if $$f=\min_{K\cap L}(\alpha)$$. Then $$\min_{K}(\alpha)$$ divides $$f$$. But if $$f$$ has repeated roots then I don't see what goes wrong.

Thanks!

$$E/F$$ is normal implies $$E/K$$ and $$E/L$$ are both normal. Since they are also separable, we get that $$E/K$$ and $$E/L$$ are Galois. Set $$G_1=\mathrm{Gal}(E/K)$$ and $$G_2=\mathrm{Gal}(E/L)$$. Note that $$E^{G_1}=K$$ and $$E^{G_2}=L$$.
Now consider $$G=\mathrm{Aut}(E/K\cap L)$$ and $$E^G$$. Obviously, $$K\cap L\leq E^G$$. Since obviously $$G_1\leq G$$ and $$G_2\leq G$$, we get that $$E^G\leq E^{G_1}=K$$ and $$E^G\leq E^{G_2}=L$$, hence $$E^G\leq K\cap L$$.
So, $$E^G=K\cap L$$, i.e. $$E^{\mathrm{Aut}(E/K\cap L)}=K\cap L$$, hence $$E/K\cap L$$ is Galois, and therefore separable.