I was struggling proving this and didn't manage to find any solution here that felt understandable for my level, so I am submitting my best idea, which seems right.
Let $L$ be a splitting field of a separable and irreducible polynomial $f\in K[t]$ over a field $K$. Show $L:K$ is a separable extension.
By the partial converse to the Fundamental Theorem of Galois Theory, I know that if the fixed field of the Galois group satisfies $Gal(L:K)^\dagger=K$, then L:K is Galois and the exercise is solved.
I also know from a Lemma on isomorphisms of simple extensions that we can extend the identity function: $id:K\to K$ to an isomorphism $\mu:K(a)\to K(b)$, $\mu\restriction_{K}=id$ and $\mu(a)=b$, where $a,b$ are two solutions of the same irreducible polynomial $f$ over $K$. By a different Lemma on isomorphisms of splitting fields, we can extend $\mu$ to an isomorphism $\lambda_{a\to b}:L\to L$, $\lambda\restriction_{K(a)}=\mu$. This would be useful since for any $a\not\in K$ I would have an element in $g\in Gal(L:K)$ where $g(a)\neq a$, so the fixed field assumption holds and $L:K$ is Galois.
Also, this procedure would justify the same result for any separable polynomial, right? It would need a little more justification since the splitting fields of the irreducible factors are not necessarily the whole $L$, but there would still be such a $g$ for every root of $f$, is what I think.
Thanks in advance!