Compactness Theorem

We prove the contrapositive: if $\Sigma$ has no model, then some finite $\Sigma_0 \subseteq \Sigma$ has no model.

Suppose $\Sigma$ is unsatisfiable. By Gödel's Completeness Theorem for first-order logic (semantic entailment coincides with syntactic derivability: $\Sigma \models \bot$ iff $\Sigma \vdash \bot$), $\Sigma$ being unsatisfiable is equivalent to $\Sigma$ being syntactically inconsistent, i.e. $\Sigma \vdash \bot$ — a contradiction is formally derivable from $\Sigma$.

A formal derivation is by definition a finite sequence of formulas, each justified by an axiom, a premise from $\Sigma$, or an inference rule applied to earlier lines. Since the derivation of $\bot$ is finite, it cites only finitely many premises from $\Sigma$; collect them into a finite set $\Sigma_0 = \{\sigma_1, \dots, \sigma_n\} \subseteq \Sigma$.

The very same finite derivation, using only premises from $\Sigma_0$, witnesses $\Sigma_0 \vdash \bot$. By soundness of first-order logic (derivability implies entailment), $\Sigma_0 \models \bot$, i.e. $\Sigma_0$ is unsatisfiable — it has no model.

So we have produced a finite $\Sigma_0 \subseteq \Sigma$ with no model, proving the contrapositive. Equivalently: if every finite subset of $\Sigma$ has a model, no such inconsistent finite derivation can exist, so $\Sigma$ cannot be unsatisfiable, hence $\Sigma$ has a model.