I'm trying to understand the following proof of this result, but I don't understand:
where the assumption that $T$ has no finite models is used;
why the final step is valid.
Take two models $\mathcal{M},\mathcal{N}$ of $T$, and suppose $\mathcal{M}\models\phi$. We need to show $\mathcal{N}\models\phi$ (the proof works exactly the same if $\mathcal{M}\not\models \phi$).
By the definition I have of $\omega$-categorical, $T$ must be in a countable language with an infinite model. So, by the Downward Lowenheim Skolem theorem, $\mathcal{M},\mathcal{N}$ both have countable elementary substructures, $\mathcal{M}',\mathcal{N}'$. Since $T$ is $\omega$-categorical, $\mathcal{M}'$ is isomorphic to $\mathcal{N}'$.
Thus far I am fine, but the proof then concludes "by elementarily, $\mathcal{N}\models\phi$". This is the step that I don't follow. What I have thought of is: "by the Tarski-Vaught test, $\mathcal{M}\models \phi$ precisely when $\mathcal{M}' \models \phi$, which is precisely when $\mathcal{N}'\models\phi$, and if $\mathcal{N}'\models\phi$ then $\mathcal{N}$ models $\phi$ since $\mathcal{N}'$ is a substructure", but I'm not comfortable with this.