I would like an explanation for the fact stated in the title. To repeat:

Question: How does one prove that if a field has a separable extension of degree $n$, then it has a Galois extension of degree $n$?

Note that the statement might as well be false (though I would not bet on that). [EDIT: I thought I had a proof that the statement was true for $n=3$, but as Emil Jeřábek points out in his comments, this is false as well!]

This might be completely trivial, but I have no idea on how to address this question, and since it kinds of naturally arose in my research, I thought it would be appropriate to ask here.

failsalready for $n=3$. Let $K/F$ be any non-normal extension of degree 3, and let $L/F$ be its normal closure. We have $[L:K]=2$. Using Zorn's lemma, let $F'$ be a maximal algebraic extension of $F$ such that $F'\cap L=F$, and let $K'$ and $L'$ be the composita of $F'$ with $K$ and $L$, respectively. We still have that $K'$ is a degree-3 extension of $F'$, and $L'$ is its normal closure over $F'$, of degree 2 over $K'$. $Gal(L'/F')\simeq S_3$. Assume for contradiction that $F'$ has a degree-3 Galois extension $H$... $\endgroup$ – Emil Jeřábek Sep 20 '18 at 14:32