[proofplan] We apply the [Downward Löwenheim-Skolem Theorem](/theorems/4272) to the expanded structure $\mathcal{H}$ with the prescribed parameter set $A$. The countability of both the language $\mathcal{L}$ and the set $A$ lets us choose the target size $\aleph_0$. The resulting elementary substructure has countable universe, contains $A$, and is a subset of $H_\theta$ because it is a substructure of $\mathcal{H}$. [/proofplan] [step:Apply downward Löwenheim-Skolem with the parameter set $A$] Let $\kappa := \aleph_0$. Since $\mathcal{L}$ is countable, we have $|\mathcal{L}| \leq \kappa$. Since $A$ is countable, we also have $|A| \leq \kappa$. The structure $\mathcal{H}$ has universe $H_\theta$, and $H_\theta$ is infinite by hypothesis. By the [Downward Löwenheim-Skolem Theorem](/theorems/4299) (citing a result not yet in the wiki: Downward Löwenheim-Skolem Theorem), applied to the infinite $\mathcal{L}$-structure $\mathcal{H}$, the set of parameters $A \subset H_\theta$, and the cardinal $\kappa = \aleph_0$, there exists an $\mathcal{L}$-substructure $\mathcal{M} \subset \mathcal{H}$ such that \begin{align*} A \subset M, \qquad \mathcal{M} \preccurlyeq \mathcal{H}, \qquad |M| \leq \kappa. \end{align*} Here $M$ denotes the universe of $\mathcal{M}$. [/step] [step:Identify the resulting substructure as the required countable subset of $H_\theta$] Because $\mathcal{M}$ is an $\mathcal{L}$-substructure of $\mathcal{H}$, its universe satisfies $M \subset H_\theta$. The inequality $|M| \leq \aleph_0$ says exactly that $M$ is countable. The construction also gives $A \subset M$ and $\mathcal{M} \preccurlyeq \mathcal{H}$. Thus there is a countable $M \subset H_\theta$ containing $A$ whose induced $\mathcal{L}$-substructure is elementary in the chosen expansion of $H_\theta$. The argument makes no claim that $M$ is transitive; it is only a countable elementary substructure of the expanded structure on $H_\theta$. [/step]