$\impliedby$: Trivial.

$\implies$: Take any non-constant $f(x) \in \Omega[x]$. By Theorem 1504, $f(x)$ has a splitting field $\Omega'$ and $\Omega' / \Omega$ is finite thus algebraic by Proposition 1404. Then by Proposition 14012, since $\Omega' / \Omega$ and $\Omega / F$ are algebraic, $\Omega' / F$ is algebraic too. Hence for any $\alpha \in \Omega'$, there exists $a(x) \in F[x]$ such that $a(\alpha) = 0$. Now, since $\Omega$ is an algebraic closure of $F$, $a(x)$ splits completely in $\Omega$, which implies $\alpha \in \Omega$. Therefore, $\Omega' = \Omega$. So every non-constant $f(x) \in \Omega[x]$ splits completely over $\Omega$, $\Omega$ is algebraically closed.