The one-step root condition makes an algebraic extension of a perfect field algebraically closed

Let $f\in F[x]$ be irreducible and nonconstant, and choose a splitting field $E\mathrm{/}F$ by [L4]. It is generated by the finitely many roots of $f$, so [L7] makes it finite; [L1] makes it separable and [L2] gives $E=F(\alpha )$ for some $\alpha$.

The minimal polynomial $m_{\alpha }\in F[x]$ has a root $\beta \in L$ by hypothesis. By [L3] there is an $F$-embedding $E=F(\alpha )\to L$ sending $\alpha$ to $\beta$. Since $f$ splits in $E$ and its coefficients are fixed, it splits in the image inside $L$.

Thus every irreducible polynomial over $F$, and hence every polynomial over $F$, splits in $L$.

Let $q\in L[x]$ be nonconstant and choose a root $\gamma$ in a splitting field by [L4]. The element $\gamma$ is algebraic over $L$, while $L\mathrm{/}F$ is algebraic, so [L5] makes $\gamma$ algebraic over $F$. Its minimal polynomial over $F$ splits in $L$ by step 3.1; since $\gamma$ is one of its roots, $\gamma \in L$.

Every nonconstant polynomial over $L$ therefore has a root in $L$, so [L6] makes $L$ algebraically closed.