Proof. Consider the diagram $$\xymatrix @R1pc@C1pc{ & \ar@{-}[ld]\ar@{-}[rd]EF \\ \ar@{-}[rd]E & & \ar@{-}[ld]F\\ & \ar@{-}[d]E\cap F \\ & K } $$ (1): Let $p(x)\in K[x]$ with a root $\alpha\in E\cap F$ Then since $E$ (resp. $F$ is Galois over $K$ all the roots of $p$ lie in $E$ (resp. $F$ and thus in $E\cap F$ The result follows.

(2): We first show that $EF$ is Galois over $K$ Choose separable polynomials $p(x),q(x)\in K[x]$ so that $E$ (resp. $F$ is a splitting field for $p$ (resp. $q$ . Then $EF$ is a splitting field for the squarefree part of $pq$ which is separable since it is squarefree and since $p(x),q(x)$ are separable.

Now, define $$\theta: \Gal(EF/K)\to \Gal(E/K)\times \Gal(F/K): \sigma\mapsto (\sigma|_E,\sigma|_F)$$ This map is a group homomorphism; its kernel is precisely those elements that leave both $E$ and $F$ fixed. Any such element must thus leave $EF$ fixed, so that $\theta$ is injective. The image obviously lies in $$H=\{ (\sigma, \tau) : \sigma|_{E\cap F}=\tau|_{E\cap F} \}$$ by construction: $(\sigma|_E)|_{E\cap F} = \sigma|_{E\cap F} = (\sigma|_F)|_{E\cap F}$ We will show that $H$ is precisely the image of $\theta$ by showing that the order of $H$ is the same as the index of the field extension $[EF:K]$

For each $\sigma\in \Gal(E/K)$ there are precisely $\Order{\Gal(F/E\cap F)}$ elements of $\Gal(F/K)$ whose restrictions to $E\cap F$ are $\sigma|_{E\cap F}$ Thus directly from the definition of $H$ $$ \Order{H} = \Order{\Gal(E/K)}\cdot\Order{\Gal(F/E\cap F)} = \Order{\Gal(E/K)}\cdot\frac{\Order{\Gal(F/K)}}{\Order{\Gal((E\cap F)/K)}} $$ By the corollary to the theorem regarding the compositum of a Galois extension and another extension, we have $$[EF:K] =[EF:F][F:K] = [E:E\cap F][F:K] = \frac{[E:K][F:K]}{[E\cap F:K]}$$ so that $$ \Order{H} = [EF:K] $$