I am a bit stuck on how I am meant to approach this.
Firstly, I think of the fact that, as $E/F$ is Galois, $|Gal(E/F)|=|E:F|=|E:K||K:F|=|E:K|2$. Then, I know that an intermediate field of a Galois extension is Galois if and only if its Galois group is normal in the larger one, that is, $Gal(K/F)\lhd Gal(E/F)$. Well, I know that any subgroup of index 2 must be normal, but it doesn't seem as if we can show that just from saying the degree is 2. Even worse, if it was index 2, I think that would require the entire extension would be degree 4, which is certainly not a requirement for the hypothesis here.
Also, I do know that any degree 2 extension is Galois as long as the field has characteristic greater than 2, but I am not confident we know this for this question.
Best Answer
You have all the right ingredients, but some misunderstanding: $\text{Gal}(K/F)$ is in general not a subgroup of $\text{Gal}(E/F)$, let alone a normal one: Pick $g\in \text{Gal}(K/F)$, how does $g$ act on elements from $E\setminus K$? However if you have $g\in \text{Gal}(E/F)$, you might be able to restrict $g|_{K}\in \text{Gal}(K/F)$ if $g$ leaves $K$ invariant, so at best (when $K/F$ is indeed Galois), $\text{Gal}(K/F)$ should be a quotient group of $\text{Gal}(E/F)$.
The fundamental theorem of Galois theory in fact considers the subgroup $$G_K:=\{g\in G:=\text{Gal}(E/F) \mid \forall x\in K, g(x)=x\}$$ and shows that $[G:G_K]=[K:F]$, and further $K/F$ is Galois iff $G_K\lhd G$ in which case $\text{Gal}(K/F)\simeq G/G_K$.
In your case, as $[K:F]=2$, $[G:G_K]=2$, so $G_K$ is normal according to group theory.
Another way to think about it is that an extension of degree $2$ is always normal, but not necessarily separable. As a subextension of a Galois extension (which is separable), $K/F$ must be separable, hence Galois.