I'm trying to prove the following result for formula $F$ and set of formulas $\mathcal{S}$:
$\mathcal{S} \models F\Rightarrow \exists \mathcal{S}_0 ⊆ S$ (finite) such that $\mathcal{S}_0 \models F$
Take $\mathcal{A}$ such that $\mathcal{A}(\mathcal{S})=1$ so that $\mathcal{A}(F)=1$. By the compactness theorem, there is finite $\mathcal{S}_0 ⊆ S$ which is also satisfiable; indeed, we can say $\mathcal{A}(\mathcal{S}_0) = 1$ by the fact that $\mathcal{S}_0 ⊆ S$.
If what I have done is correct (I'm not sure) then I believe what we are looking for is to ensure that all other assignments such that $\mathcal{A}(\mathcal{S}_0) = 1$ also give $\mathcal{A}(F) = 1$. However, I'm really not sure how to do this and was wondering if anyone could help? Thanks!