Compositum of fields with trivial intersection
This is false in general. Let me begin by offering an example from the side of finite groups (the Galois correspondence is inclusion reversing, so...). Let $G=D_8$ be the group of symmetries of a regular octagon. We know that $G$ is generated by a rotation $r$ of order $8$, and a reflection $s$ of order $2$ such that $srs=r^{-1}$, and it follows that $sr^i=r^{8-i}s$ for all integers $i$. It follows that $|G|=16$ and that $r^4$ is the only non-trivial element in the center of $G$. Geometrically this is obvious, because $r^4$ is the rotation by the angle $\pi$, i.e. scalar multiplication by $-1$.
Inside $G$ we have two subgroups $H_1=\{1,s,r^4, sr^4\}$ and $H_2=\{1,sr,r^4,sr^5\}$. Both of these are generated by reflections w.r.t. to a pair of orthogonal lines - one pair gotten by rotating the other by the angle $\pi/8$. Geometrically: one pair of the the orthogonal lines connects two pairs of antipodal vertices of the octagon, the other pair connects antipodal midpoints of edges. Anyway, the key observations are the following:
- $\langle H_1,H_2\rangle=G$. This is because $r=s(sr)$ is in the subgroup generated by both $H_1$ and $H_2$. In other words no proper subgroup of $G$ contains both $H_1$ and $H_2$.
- $H_1\cap H_2=\{1,r^4\}$ is non-trivial.
Translating this to field extensions forming a counterexample is straightforward. Let $E/F$ be a Galois extension with Galois group $G$. Let $K=\operatorname{Inv}(H_1)$ and $L=\operatorname{Inv}(H_2)$. The intersection $L\cap K$ is then invariant under $H_1$ and $H_2$, hence under all of $G$, so $L\cap K=F$. OTOH the compositum $LK$ is the smallest field containing both $L$ and $K$, and thus the field of invariants of $H_1\cap H_2$. Here $[K:F]=[G:H_1]=4$, $[L:F]=[G:H_2]=4$, $[E:F]=|G|=16=4\cdot4$, but $[LK:F]=[G:(H_1\cap H_2)]=8<16$.
Translating everything to the group side by an application of the Galois correspondence was the key. After that it was just tinkering, trial and error.
Observe that it suffices to find a finite group $G$ together with two subgroups $H_1$ and $H_2$ such that i) $G=\langle H_1, H_2\rangle$, ii) $|G|=[G:H_1]\cdot[G:H_2]$ and ii) $|H_1\cap H_2|>1$.