Let $F$ be a perfect field, i.e. every irreducible polynomial over $F$ has distinct roots in the algebraic closure of $F$. Suppose that $K$ is an algebraic extension of $F$ with the property that every non-constant $p(X) \in F[X]$ has a root in $K$.
I want to show that $K$ is algebraically closed, i.e. that $K = \overline{F}$, the algebraic closure of $F$.
If $K$ were a normal extension of $F$, this would follow immediately: Let $\alpha \in \overline{F}$ and $m(X)\in F[X]$ its minimal polynomial. By our assumption, $m(X)$ has at least one zero in $K$. Since $K$ is a normal extension of $F$, it thus contains all other roots of $m(X)$ including $\alpha$.
I don't have an idea as to how to show that $K$ is a normal extension.
The trick is to use the primitive element theorem. For any $f\in F[x]$, let $L$ be a splitting field of $f$. Then $L$ is finite over $F$ and separable (since $F$ is perfect), so it has a primitive element $\beta$. We then see that $f$ splits over any extension of $F$ that has a root of the minimal polynomial of $\beta$. In particular, $f$ splits over $K$, as desired.