Finitely Satisfiable Theory has Maximal Finitely Satisfiable Extension/Lemma

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

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

Then either $T \cup \{ \phi \}$ or $T \cup \{ \neg \phi \}$ is finitely satisfiable.

Proof
Suppose WLOG that $\phi \notin T$.

Suppose that $T \cup \{\phi\}$ is not finitely satisfiable.

Then by definition there must be a finite subset $K$ of $T \cup \{\phi\}$ which is not satisfiable.

Since $T$ is finitely satisfiable, $\phi \in K$, so $\Delta = K \setminus \{ \phi \}$ is a finite subset 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 either in $T$ or of the form $\Sigma \cup \left\{{\neg \phi}\right\}$.