$F/K$ 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?

Let $E/F$ be a finite extension (which is separable as the characteristic is zero), we will prove that $E=F$, hence $F$  is algebraically closed:
Let $a$ be a primitive element of $E/F$. Then $a$ is algebraic over $K$, so there exists a finite Galois extension $N/K$  containing $a$, in particular 
$$
E\subseteq NF.
$$
(One can take $N$ to be the Galois closure of $K(a)$, for example.)
Let $f\in K[x]$ be the minimal polynomial of a primitive element of $N$ over $K$. Since $N/K$ is Galois, any of the roots of $f$ generates $N$ over $K$. 
Now, by assumption $f$ has a root in $F$, so $N\subseteq F$, hence $E\subseteq NF=F$. This proves that $F$ has no non-trivial extensions, hence $F$ is algebraically closed. 
