The argument doesn't actually require an algebraic closure; you can just replace the algebraic closure by sufficiently large finite extensions in each step of the argument.
Specifically, say $F$ is the splitting field of $f$ over $K$ and let $L$ be an extension of $F$ over which $g$ splits. Let $\alpha\in F$ be a root of $g$ and let $\beta\in L$ be another root of $g$; we wish to show $\beta\in F$. There is an embedding $i:K(\alpha)\to L$ which fixes $K$ and sends $\alpha$ to $\beta$. This embedding can then be extended to an embedding $j:F\to L'$ for some finite extension $L'$ of $L$ (in fact, you can prove that you can take $L'=L$, but that is not needed for this argument). But then both $j(F)$ and $F$ are splitting fields of $f$ over $K$ inside $L'$, so $j(F)=F$. Since $j(\alpha)=\beta$, this implies $\beta\in F$, as desired.
(In case you are skeptical of the existence of $L'$, here is the general lemma I am using. If $i:k\to L$ is a field embedding and $F$ is a finite extension of $k$, then there exists a finite extension $L'$ of $L$ and an embedding $j:F\to L'$ extending $i$. Indeed, $F$ is generated over $k$ by finitely many elements, and so by extending one element at a time, we may assume $F=k(a)$ for a single element $a$. Now let $h$ be the minimal polynomial of $a$ over $k$ and let $L'$ be an extension of $L$ obtained by adjoining a root $b$ of the polynomial obtained by applying $i$ to all the coefficients of $h$. There is then a unique extension of $i$ to an embedding $j:F\to L'$ which sends $a$ to $b$.)