Upward Löwenheim-Skolem Theorem

Theorem
Let $T$ be an $\mathcal{L}$-theory with an infinite models. Then for each infinite cardinal $\kappa\geq |\mathcal{L}|$, there is a model of $T$ with cardinality $\kappa$.

Proof
The idea is to extend the language by adding $\kappa$ many new constants  and extend the theory by adding sentences asserting that these constants  are distinct. We can show that this new theory is finitely satisfiable using an infinite model of $T$. Compactness then implies that the new theory has a model. Some care needs to be taken to ensure that we construct a model of exactly size $\kappa$.

Let $\mathcal{L}^*$ be the language formed by adding new constants  $\{c_\alpha \mid \alpha < \kappa \}$ to $\mathcal{L}$. Let $T^*$ be the $\mathcal{L}^*$-theory formed by adding the sentences $\{c_\alpha  \neq c_\beta \mid \alpha, \beta < \kappa,\ \ \alpha\neq\beta\}$ to  $T$.

We show that $T^*$ is finitely satisfiable:

Let $\Delta$ be a finite subset of $T^*$. Then $\Delta$ contains finitely many sentences from $T$ along with finitely many sentences of the form  $c_\alpha \neq c_\beta$ for the new constant symbols. Since $T$ has an infinite model, it must have a model $\mathcal{M}$ of cardinality at  most $|\mathcal{L}|+\aleph_0$. This model already satisfies everything in $T$. So, since we can find arbitrarily many distinct elements in it, it can also be used as a model of $\Delta$ by interpreting the finitely  many new constant symbols in $\Delta$ as distinct elements of  $\mathcal{M}$.

Since $T^*$ is finitely satisfiable, it follows by the Compactness Theorem that $T^*$ itself is satisfiable. Since $T^*$ ensures the existence of $\kappa$ many distinct elements, this means it has models of size at least $\kappa$. It can be proved separately or observed from the ultraproduct proof of the compactness  theorem that $T^*$ then has a model $\mathcal{M}^*$ of exactly size  $\kappa$.

Since $T^*$ contains $T$, $\mathcal{M}^*$ is a model of $T$ of size $\kappa$.