[proofplan]
The argument is a direct compactness contradiction. The finite satisfiability hypothesis says that every finite fragment of $\Gamma \cup \Sigma$ has a model, so first-order compactness produces a model of the full theory $\Gamma \cup \Sigma$. Since this model satisfies every sentence of $\Gamma$, the hypothesis that every model of $\Gamma$ belongs to $\mathcal{C}$ places it inside $\mathcal{C}$, contradicting the hypothesis that no member of $\mathcal{C}$ satisfies $\Gamma \cup \Sigma$.
[/proofplan]
[step:Apply compactness to the finitely satisfiable combined theory]
Let $\Theta$ denote the set of first-order $\mathcal{L}$-sentences
\begin{align*}
\Theta := \Gamma \cup \Sigma.
\end{align*}
This is a first-order $\mathcal{L}$-theory because $\Gamma$ is a first-order $\mathcal{L}$-theory and $\Sigma$ is a set of first-order $\mathcal{L}$-sentences. By hypothesis, every finite subset of $\Theta$ has an $\mathcal{L}$-model. Therefore, by the First-Order [Compactness Theorem](/theorems/2748), the full theory $\Theta$ has an $\mathcal{L}$-model. Thus there exists an $\mathcal{L}$-structure $M$ such that
\begin{align*}
M \models \Gamma \cup \Sigma.
\end{align*}
[guided]
The purpose of introducing $\Theta$ is only to give a name to the combined theory whose finite fragments are assumed satisfiable. Define
\begin{align*}
\Theta := \Gamma \cup \Sigma.
\end{align*}
Because $\Gamma$ is a first-order $\mathcal{L}$-theory and $\Sigma$ is a set of first-order $\mathcal{L}$-sentences, the union $\Theta$ is itself a set of first-order $\mathcal{L}$-sentences. The hypothesis says exactly that for every finite set $\Theta_0 \subset \Theta$, there is an $\mathcal{L}$-structure $M_{\Theta_0}$ such that
\begin{align*}
M_{\Theta_0} \models \Theta_0.
\end{align*}
This is the hypothesis required by the First-Order Compactness Theorem: if every finite subset of a first-order theory has a model, then the entire theory has a model. Applying that theorem gives an $\mathcal{L}$-structure $M$ satisfying every sentence in $\Theta$. Since $\Theta = \Gamma \cup \Sigma$, this means
\begin{align*}
M \models \Gamma \cup \Sigma.
\end{align*}
[/guided]
[/step]
[step:Use the proposed axiomatization to place the compactness model inside $\mathcal{C}$]
Since $M \models \Gamma$ and every $\mathcal{L}$-structure satisfying $\Gamma$ belongs to $\mathcal{C}$, we have
\begin{align*}
M \in \mathcal{C}.
\end{align*}
But the same structure also satisfies $M \models \Gamma \cup \Sigma$, contradicting the hypothesis that no $\mathcal{L}$-structure in $\mathcal{C}$ satisfies $\Gamma \cup \Sigma$.
[guided]
The compactness model $M$ satisfies all sentences in $\Gamma \cup \Sigma$. In particular, it satisfies every sentence in $\Gamma$:
\begin{align*}
M \models \Gamma.
\end{align*}
The assumed defining property of $\Gamma$ is that every model of $\Gamma$ lies in the class $\mathcal{C}$. Applying that assumption to the structure $M$ gives
\begin{align*}
M \in \mathcal{C}.
\end{align*}
At the same time, compactness gave
\begin{align*}
M \models \Gamma \cup \Sigma.
\end{align*}
This is impossible under the second hypothesis on $\Sigma$, namely that no $\mathcal{L}$-structure belonging to $\mathcal{C}$ satisfies $\Gamma \cup \Sigma$. Thus the assumptions force a contradiction.
[/guided]
[/step]
[step:Conclude that the finite satisfiability obstruction rules out the proposed definition]
The contradiction shows that no first-order $\mathcal{L}$-theory $\Gamma$ can simultaneously satisfy the stated hypotheses. In particular, a theory that axiomatizes $\mathcal{C}$ cannot satisfy them, because axiomatizing $\mathcal{C}$ includes the condition that every model of $\Gamma$ belongs to $\mathcal{C}$. If a single sentence $\varphi$ is claimed to define $\mathcal{C}$, then the same argument applies to the theory $\Gamma = \{\varphi\}$, so $\varphi$ cannot define $\mathcal{C}$ under this finite satisfiability obstruction.
[guided]
We have derived a contradiction from the simultaneous assumptions that every model of $\Gamma$ belongs to $\mathcal{C}$, that every finite subset of $\Gamma \cup \Sigma$ has an $\mathcal{L}$-model, and that no $\mathcal{L}$-structure in $\mathcal{C}$ satisfies $\Gamma \cup \Sigma$. Therefore no first-order $\mathcal{L}$-theory $\Gamma$ can satisfy all of these hypotheses. This proves the advertised non-definability criterion: if $\Gamma$ were an axiomatization of $\mathcal{C}$, then every model of $\Gamma$ would belong to $\mathcal{C}$, so the same contradiction would rule out such a $\Gamma$ whenever the finite satisfiability obstruction is present.
For the final sentence of the theorem, suppose a single first-order $\mathcal{L}$-sentence $\varphi$ is proposed as a definition of $\mathcal{C}$. Define the first-order $\mathcal{L}$-theory $\Gamma$ by
\begin{align*}
\Gamma := \{\varphi\}.
\end{align*}
Then the preceding contradiction applies with this choice of $\Gamma$. Hence the sentence $\varphi$ cannot define $\mathcal{C}$ under the stated finite satisfiability obstruction.
[/guided]
[/step]
