Finitely Satisfiable Theory has Maximal Finitely Satisfiable Extension/Proof 2

Theorem
Let $T$ be a finitely satisfiable $\LL$-theory.

There is a finitely satisfiable $\LL$-theory $T'$ which contains $T$ as a subset such that for all $\LL$-sentences $\phi$, either $\phi \in T'$ or $\neg\phi \in T'$.

Proof
Let $\AA$ be the set of finitely satisfiable extensions of $T$.

By the lemma, for each element $S$ of $\AA$ and each $\LL$-sentence $\phi$, either $S \cup \set \phi \in \AA$ or $S \cup \set {\neg \phi} \in \AA$.

$\AA$ has finite character, by the following argument:

Let $S \in \AA$.

Let $F$ be a finite subset of $S$.

Then $S$ is satisfiable and hence finitely satisfiable.

Thus in $\AA$.

Let $S$ be a theory on $\LL$.

Let every finite subset of $S$ be finitely satisfiable.

Then every finite subset of $S$ is satisfiable.

Therefore $S$ is finitely satisfiable.

Thus $\AA$ has finite character.

By the Restricted Tukey's Theorem, $\AA$ has an element $T'$ such that:
 * for each $\LL$-sentence $\phi$, either $\phi \in T'$ or $\neg \phi \in T'$.