One can check that $F(S)$ is a splitting field for $A$ over $F$, where $S \subseteq \overline{F}$ is the subset consisting of roots of the polynomials in $A$. (cf. Proposition 1503)
It suffices to show that if $K$ is a splitting field for $A$ over $F$, then there exists a field isomorphism $\varphi : K \to F(S)$ with $\varphi |_{F} = \operatorname{id}_{F}$. Apply Proposition 1603 tothen we obtain a field embedding $\varphi : K \to \overline{F}$. Every polynomial splits completely over $K$ $\implies$ $S \subseteq \varphi(K)$ $\implies$ $F(S) \subseteq \varphi(K)$ $\implies$ $\varphi ^{-1}(F(S)) \subseteq K$. By (2) of Definition 2202, since $K$ is a splitting field for $A$ over $F$, it follows that $\varphi ^{-1}(F(S)) = K$ $\implies$ $F(S) = \varphi(K)$. Moreover, $\varphi$ is injective and $\varphi |_{F} = \operatorname{id}_{F}$. So $\varphi$ is a field isomorphism that maps $K$ to $F(S)$ with $\varphi |_{F} = \operatorname{id}_{F}$.
