This question already has an answer here:
Let $F\subset E$ be an algebraic extension of fields and suppose that every polynomial $f(x) \in F[x]$ has a root in $E$. Is it possible that $E$ is not algebraically closed?
Note that such an example must necessarily be of prime characteristic since over char $0$, $E$ is necessarily algebraically closed. Proof:
Let $g(x) \in E[x]$ be a polynomial with a root $\alpha$ (after fixing an algebraic closure of $E$). Let $f(x)$ be the minimal polynomial of $\alpha$ over $F$ and let $L$ be the splitting field of $f$. By the primitive element theorem, $L=F[\beta]$ and if $h(x)$ is the minimal polynomial for $\beta$, it has some root $\gamma \in E$. Since $F[\beta] = F[\gamma]$, this shows that $L\subset E$.
In particular, if we are looking for counterexamples, we would like $L$ to not be a primitive extension.