Intersection of finite Galois extensions is Galois
Solution 1:
If we are taking their intersection, then presumably we are viewing them as subfields of some unmentioned field $M$. Without loss of generality $M$ is separably closed, making it the separable closure of $F$ (have you heard of this?). The separable closure is an algebraic separable extension of $K$ which has no algebraic separable extensions except for itself. Now
A subfield of a separable extension is separable. It follows that $L \cap K$ is separable.
Each $F$-automorphism $\sigma$ of $K$ has $\sigma(L) = L$ and $\sigma(K) = K$, so $\sigma(L \cap K) \subset L \cap K$. It follows that $\sigma(L \cap K) = L \cap K$ by degree considerations (though really we didn't need that $L$ and $K$ were finite). This shows that $L \cap K$ is normal.
Here is a different perspective. While it may be harder at first, internalizing it makes results like these easier to see:
Let $F$ be a field. Instead of a fundamental theorem of galois theory for each extension $K/F$, we have a fundamental theorem of galois theory for the largest Galois extension $F^{sep}/ F$. $F^{sep}$ here is the mentioned "separable closure", defined up to isomoprhism just like the algebraic closure. Here are some descriptions of it:
$F^{sep}$ can be constructed as the set of element in the algebraic closure of $F$ whose minimal polynomial in $F[x]$ splits into distinct factors in the algebraic closure.
$F^{sep}$ is separable algebraic (not necessarily finite) and any separable algebraic extension of $F^{sep}$ is identically $F^{sep}$.
$F^{sep}$ is separable algebraic (not necessarily finite separable) and any finite separable extension of $F^{sep}$ is identically $F^{sep}$.
It turns out that the extension $F^{sep} / F$ is Galois, and in a sense it is the largest Galois extension of $F$ (note it is not unique up to unique isomorphism). It has a galois group $\text{Aut}_F (F^{sep})$, called the absolute galois group, which has a topology on it (that part is a little involved, so I'll spare the details).
What do we know about subfields of Galois extensions? They correspond to subgroups of the Galois group! Any separable field extension $K$ of $F$ can be viewed as a subfield of $F^{sep}$, and corresponds to a subgroup of the absolute galois group (which is also open in the topology).
To recover the ordinary fundamental theorem of galois theory, finite galois field extensions $K$ of $F$ correspond to certain normal subgroups $N$ of the absolute galois group $G$ of $F$. Subfields of such a Galois extension $K$ of $F$ will correspond to subgroups of the absolute galois group $G$ containing $N$. Then we have a correspondence: $$ \{ L \subset K: L \text{ a subfield} \} \leftrightarrow \{ N \subset H \subset G : H \text { an open subgroup} \} \leftrightarrow \{ \overline{H} \subset G / N : \overline{H} \text{ an open subgroup} \} $$ $G/N$ turns out to be the galois group of $K/F$. So really we have packed all the Galois groups into one big one- all finite Galois groups of a galois extension of $F$ occur as quotients of the biggest one.
Up to isomorphism, finite separable extensions of $F$ correspond to open subgroups of the absolute galois group, and galois extensions of $F$ correspond to open normal subgroups of the absolute galois group.
I mentioned this perspective makes the theorem obvious. The reason is that intersections of fields in the separable closure correspond to joins of subgroups in the absolute galois group. From this vantage point, it is automatic that the intersection of fields in the separable closure is separable (any subfield of the separable closure is separable, and all separable extensions occur in the separable closure up to isomorphism). That they are normal corresponds to the fact that the group generated by normal subgroups is normal, and that is not so hard to show.