Combining the answers of Joel and Kevin does work.

Write $\overline{\mathbb{Q}} = \bigcup_{n=0}^\infty K_n$ where each $K_n$ is a finite Galois extension of $\mathbb{Q}$. Let $S_n$ be the primes that split completely in $K_n$. By Chebotarev's Theorem, this is a descending sequence of infinite sets of primes. Therefore, there is a nonprincipal ultrafilter $\mathcal{U}$ over the set of primes which contains all $S_n$. If $f(x) \in \mathbb{Z}[x]$ has a root in $K_n$, then $f(x)$ has roots in $\mathbb{F}_p$ for all $p \in S_n$ and hence $f(x)$ has a root in the ultraproduct $\prod_p \mathbb{F}_p/\mathcal{U}$.

Combining the answers of Joel and Kevin does work.

Write $\overline{\mathbb{Q}} = \bigcup_{n=0}^\infty K_n$ where the $K_n$ are an increasing sequence of finite Galois extension of $\mathbb{Q}$. Let $S_n$ be the primes that split completely in $K_n$. By Chebotarev's Theorem, this is a descending sequence of infinite sets of primes. Therefore, there is a nonprincipal ultrafilter $\mathcal{U}$ over the set of primes which contains all $S_n$. If $f(x) \in \mathbb{Z}[x]$ has a root in $K_n$, then $f(x)$ has roots in $\mathbb{F}_p$ for all $p \in S_n$ and hence $f(x)$ has a root in the ultraproduct $\prod_p \mathbb{F}_p/\mathcal{U}$.