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 $T \cup \left\{ {\phi}\right\}$ or $T \cup \left\{ {\neg \phi}\right\}$ is finitely satisfiable.

Proof
, suppose that $\phi \notin T$.

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

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

Since $T$ is finitely satisfiable, $\phi \in K$.

Therefore $\Delta = K \setminus \left\{ {\phi}\right\}$ 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 $\left\{ {\neg \phi}\right\} \cup \Sigma$ is satisfiable.

But any finite subset of $T \cup \left\{ {\neg \phi}\right\}$ is either in $T$ or of the form $\Sigma \cup \left\{{\neg \phi}\right\}$.

This demonstrates that $T \cup \left\{ {\neg \phi}\right\}$ is finitely satisfiable.