The following statement appears in Corollary 4.10 of Morandi's book titled Field and Galois Theory
Let L be a finite extension of F. Then, L is separable over F if and only if it is contained in a Galois extension of F.
The result is proved in the text assuming finiteness of the extension $L/F$. I believe that the above statement still holds if the condition $L/F$ is finite is relaxed to $L/F$ is algebraic. Here's my argument:
Suppose that $L$ lies in $K$ for some Galois extension $K$ of $F$, then $K/F$ is separable. Hence, $L/F$ is separable as well. Conversely, suppose that $L/F$ is separable. Then, for every $l\in L$, we know that $l$ satisfies a separable minimal polynomial, say $l_F$ over $F$. Suppose that $l_F$ factors in $L[X]$ as $f=(x-l_1)(x-l_2)\ldots (x-l_n)f_{(1,l)}\ldots f_{(r_l,l)}$, where $l_1,\ldots , l_{n_l}\in L$ and $f_{(1,l)}, \ldots ,f_{(r_l,l)}\in L[X]$ are irreducible with degree at least $2$. Let $K$ be a splitting field over $L$ of the collection of polynomials $\{f_{(1,l)},\ldots,f_{(r_l,l)}|l\in L\}$. Then, by definition $K$ is a splitting field of the family $\{l_F|l\in L\}$ of separable polynomials over $F$. Hence, $K/F$ is Galois and $L\subseteq K$. $\;\;\;\;\;\;\;\;\;\;\blacksquare$
I would be grateful if the above claim and its proof can be verified. If I'm mistaken, please provide details on where I'm going wrong. Thank you!