Apply Proposition 1603 to $L / F$ and the natural field embedding $F \to L'$, then there exists a field embedding $\varphi : L \to L'$ such that $\varphi |_{F} = \operatorname{id}_{F}$. By Proposition 1605, $L$ and $L'$ are both algebraically closed. Then $L$ is algebraically closed $\implies$ $\varphi(L)$ is algebraically closed since $L \cong \varphi(L)$ (recall that a field homomorphism is either $0$ or its kernel is $0$). By Proposition 14012, $L' / F$ is algebraic $\implies$ $L' / \varphi(L)$ is algebraic. By part 4 of Proposition 1601, $L' = \varphi(L)$ since $L'$ is algebraically closed.