Let $\mathcal L$ be the language of rings, $T$ the field axioms, $X = \{\chi_n:n\in\Bbb N\}$ where $\chi_n$ expresses in $\mathcal L$ that every degree $n$ polynomial has a root. Let $F$ be any field and define a set of constant symbols $C=\{c_x:x\in F\}$ such that $\mathcal L\cap C = \emptyset$. Extend $\mathcal L$ to the language $\mathcal L' = \mathcal L\cup C$. It may be worth noting that this construction now gives $\mathcal L'$ two different symbols for 0 and 1, but this should end of up being consistent with everything else we do here.
Define $E = \{\eta_{i,j,k}:i,j,k\in F\}$ where $\eta_{i,j,k}$ expresses all of the additive and multiplicative relationships between $i,j,k\in F$ which can be stated with one operation followed by one equal sign. So for example, if $F=\Bbb C$ and if $i=1,j=\sqrt{-1},k=1-\sqrt{-1}$ then $\eta_{i,j,k}$ is the formula $$ c_i-c_j=c_k \land c_i\cdot c_j=c_j \land ...$$
I am trying to prove that every finite subset of $T\cup X\cup E$ has a model.
Initially I thought that I would try to show that $F$ can always be extended to a field $K$ which has a root for every degree $n$ polynomial, where $n$ is the largest index on any formula in the finite subset of $X$. However, I couldn't see my way through an argument that this extension always exists.
I might be able to appeal to say that for any field $F$ and polynomial $p$, the ring $F[x]/(p)$ is a field with an isomorphic copy of $F$ inside -- although I'm not sure this is true. And this is in the context of an introductory mathematical logic text that doesn't assume much background in abstract algebra, so I'd try to avoid this approach if I can. Also, even if this solves the problem for only one polynomial at a time, it's not clear that this justifies it for all of the infinitely many polynomials that might exist, of degree $n$.
I considered doing maybe some kind of proof by induction on the degrees of polynomials, but I didn't see a good path forward that way either.
Next I thought, perhaps I shouldn't just start from $F$ and extend it. Perhaps I could start from $\Bbb C$ and argue that for any finite subset of $T\cup X\cup E$ there is a subset satisfying the equations that come from $E$. However, if $F=\Bbb Z_2$ that won't be true since we could have the sentence $1+1=0$ in $E$. Together with the field axioms ensuring that $1$ is the multiplicative identity of $\Bbb C$, then we would not have $\Bbb C$ being a model of this finite subset of $T\cup X\cup E$.
Any insight would be appreciated!