Finitely Satisfiable Theory has Maximal Finitely Satisfiable Extension

Theorem
Let $T$ be a finitely satisfiable $\mathcal{L}$-theory.

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

Proof
The set of all finitely satisfiable $\mathcal{L}$-theories containing $T$ forms a partially ordered set using subset inclusion as the ordering.

Let $C$ be a nonempty chain in this partially ordered set, and let $T_C = \displaystyle \bigcup_{\Sigma \in C} \Sigma$.

If $\Delta$ is a finite subset of $T_C$, then there is a single $\Sigma$ in $C$ which contains $\Delta$. Since this $\Sigma$ is finitely satisfiable by definition, this means that $\Delta$ is satisfiable.

Hence $T_C$ is finitely satisfiable.

Since each $\Sigma \in C$ is contained in $T_C$, this means that $T_C$ is an upper bound for $C$ in the partially ordered set.

Thus, by Zorn's Lemma, there is a finitely satisfiable $\mathcal{L}$-theory $T'$ containing $T$ such that $T'$ contains all other such theories.

Let $\phi$ be an $\mathcal{L}$-sentence.

Suppose $T'$ does not contain $\phi$.

If $T' \cup \{\phi\}$ were finitely satisfiable, then by definition of  $T'$, $T'$ would contain $T' \cup \{\phi\}$ as a subset. This would mean that $T'$ contains $\phi$, which contradicts the assumption.

Thus $T' \cup \{\phi\}$ is not finitely satisfiable.

Since $T'$ is finitely satisfiable, there must be a finite subset $\Delta$ of $T'$ such that $\Delta\models \neg\phi$. Let $\Sigma$ be a finite subset of $T'$.

Since $\Delta\cup\Sigma$ is a finite subset of $T'$, it is satisfiable.

But, $\Delta\models \neg\phi$.

Hence, $\{\neg\phi\}\cup\Sigma$ is satisfiable.

This demonstrates that $T'\cup \{\neg\phi\}$ is finitely satisfiable, since  any finite subset of it is of the form $\Sigma \cup \{\neg\phi\}$.

Thus, by definition of $T'$, $T'$ contains $T'\cup\{\neg\phi\}$ as a subset, and hence $T'$ contains $\neg\phi$.