Let $\Gamma$ be a theory that is closed under provability. That is, if there are sentences $\varphi_{1},...,\varphi_{n}$ in $\Gamma$ such that $\varphi_{1},...,\varphi_{n} \vdash \phi$ it applies that $\phi \in \Gamma$. Prove that $\Gamma$ has a model if and only if each set in $\Gamma$ has a model.
I want to prove the above statement and the only theorem I can think of to prove this is the compactness theorem, that a set of sentences $\Gamma$ where every finite subset of $\Gamma$ has a model, then there is a model for the whole $\Gamma$.
But this statement deals not only with finite sets in $\Gamma$ but all sentences, infinite and finite. But my question is, can I still use the compactness theorem to prove this statement? I reason that if all sets have a model, that means that all finite sets also has a model, which results in $\Gamma$ having a model?
Also, do I need to in any way take into account that the theory is closed under provability when I do this?