Compactness Theorem/Proof using Gödel's Completeness Theorem

Proof
By definition, $T$ is finitely satisfiable means that every finite subset of $T$ is satisfiable.

Because the direction:
 * $T$ satisfiable implies $T$ finitely satisfiable

is trivial, the proof below justifies the converse:
 * $T$ finitely satisfiable implies $T$ satisfiable.

This proof is by contraposition.

The idea is to exploit the finiteness of proofs and the relation between satisfiability and deducibility to show that if $T$ is not satisfiable, then it must have a finite subset which can be used to prove to a contradiction.

Suppose $T$ is not satisfiable.

Since $T$ has no models, it vacuously follows that $T \models \phi \wedge \neg \phi$ for some sentence $\phi$.

By Gödel's Completeness Theorem, this implies that $\phi \wedge \neg \phi$ is deducible from $T$.

But any deduction from $T$ involves only finitely many sentences from $T$.

This means that there is a finite subset $\Delta$ of $T$ such that $\phi \wedge \neg \phi$ is deducible from $\Delta$.

By Soundness of First-Order Logic, this means that $\Delta \models \phi \wedge \neg \phi$.

Thus $\Delta$ is not satisfiable.

By the Rule of Transposition, $\Delta$ is satisfiable implies $T$ is satisfiable.