= Solution
\b[The printed inclusion-and-elementarity assertion is false if transitivity is required of the same submodel.] Take the theorem of <ZFC> that combines the <axiom of infinity> and the <Axiom of power set>. Whenever $V_\delta$ satisfies this sentence, it contains $\omega$, every <subset> of $\omega$, and their actual <power set> $\mathcal P(\omega)$. A submodel $M\prec V_\delta$ contains $\omega$ and $\mathcal P(\omega)$, since these are uniquely definable in $V_\delta$. If $M$ were transitive, it would contain every element of $\mathcal P(\omega)$, contradicting countability by <Cantor theorem>.
The corrected conclusion uses an <elementary embedding> rather than elementary inclusion. By <Lévy reflection theorem>, choose $\delta>\omega+2$ with $V_\delta\models\varphi$ and with Extensionality true there. The <Downward Lowenheim-Skolem theorem> says that an infinite structure in a countable language has a countable <elementary substructure>. Apply it to obtain a countable $N\prec(V_\delta,\in)$. The membership relation on $N$ is externally well-founded, and elementarity makes it extensional. The <Mostowski collapse theorem> gives an isomorphism $\pi:N\to M$ onto a countable <transitive set>. Hence
$$
\boxed{M\models\varphi,\qquad j=\pi^{-1}:M\longrightarrow V_\delta
\text{ is elementary}.}
$$
Indeed $M\subseteq V_\delta$: rank induction gives $\operatorname{rank}(\pi(x))\le\operatorname{rank}(x)$ for every $x\in N$. The crucial correction is that \b[$j$, rather than the inclusion of $M$, is elementary]. For a formula with free variables, apply this argument to its universal closure.
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