Proof. Define the propositions $$ \Phi_n \equiv \underline{\exists x_1 \cdots \exists x_n . (x_1\ne x_2)\wedge \cdots \wedge(x_1\ne x_n) \wedge(x_2\ne x_3)\wedge \cdots \wedge(x_{n-1}\ne x_n)} $$ ($\Phi_n$ says ``there exist (at least) $n$ different elements in the world''). Note that $$\cdots \vdash \Phi_n \vdash \cdots \vdash \Phi_2 \vdash \Phi_1.$$ Define a new theory $$ \T_\infty = \T \cup \left\{\Phi_1, \Phi_2, \ldots \right\}. $$ For any finite subset $\T'\subset \T_\infty$ , we claim that $\T'$ is consistent: Indeed, $\T'$ contains axioms of $\T$ , along with finitely many of $\left\{\Phi_n\right\}_{n\ge 1}$ . Let $\Phi_m$ correspond to the largest index appearing in $\T'$ . If $\M_m\models\T$ is a model of $\T$ with at least $m$ elements (and by hypothesis, such as model exists), then $\M_m\models \T\cup\{\Phi_m\}\vdash\T'$ .

So every finite subset of $\T_\infty$ is consistent; by the compactness theorem for first-order logic, $\T_\infty$ is consistent, and by Gödel's completeness theorem for first-order logic it has a model $\M$ . Then $\M\models\T_\infty\vdash\T$ , so $\M$ is a model of $\T$ with infinitely many elements ($\M\models \Phi_n$ for any $n$ , so $\M$ has at least $\ge n$ elements for all $n$ ).