$F/K$ algebraic and every nonconstant polynomial in $K[X]$ has a root in $F$ implies $F$ is algebraically closed. Let $F/K$ be an algebraic extension of fields in characteristic zero.  If $F/K$ is normal, and every nonconstant polynomial $f \in K[X]$ has a root in $F$, then $F$ is algebraically closed.  This is obvious, because if $u$ is algebraic over $F$ (let us agree to fix some algebraic closure $\overline{K}$ of $K$ containing $F$ with $u \in \overline{K}$), then it is algebraic over $K$ with minimal polynomial $f \in K[X]$, which by hypothesis must have one and therefore all of its roots in $F$.  In particular $u \in F$.
I have heard that the above assertion remains true if we drop the assumption that $F/K$ is normal.  But it seems to be a nontrivial result.  Can anyone get me started on how to prove this?
Thoughts so far: It's enough to show that every algebraic extension of $K$ is $K$-isomorphic to a subfield of $F$.  Certainly this is true for every finite extension $E$ of $K$; we're in characteristic zero, so $E = K(v)$ for some $v \in E$.  If $f \in K[X]$ is the minimal polynomial of $v$ over $K$, then $f$ has some root $u \in F$, whence $E$ is $K$-isomorphic to the subfield $K(u)$ of $F$.
So, I'm thinking next use some kind of Zorn's Lemma argument?

The question asked is more specific than the title: in the question you say $K$ has characteristic $0$, but the title is a broader question. The answer to the broader question is also "yes". That is, even when $K$ has characteristic $p$ the field $F$ is algebraically closed. This is a theorem of Robert Gilmer. See here or Theorem 2 here.
