I'm turning my comment into an answer. With $\Sigma^1_2$ statements we can discuss well-foundedness: A real codes a well-founded model of enough set theory iff it codes a model (which is an arithmetic statement) and the model is well-founded (this you can express by saying that no sequence through the ordinals of the model is strictly decreasing). So, to say that there is such a well-founded model is $\Sigma^1_2$. As for "enough set theory", pick $T$ a finite and sufficiently strong fragment of $\mathsf{ZF}$.

By the reflection theorem, there are transitive models of $T$, so there are countable transitive ones (by Lowenheim-Skolem and Mostowski). Pick an example $M$ of smallest height. We have that, in $M$, there are no transitive set models of $T$. That is, the $\Sigma^1_2$ statement discussed in the previous paragraph is true in $V$ (and in set models of certain stronger fragments of $\mathsf{ZF}$), but fails in $M$.

If we assume that there are transitive set models of $\mathsf{ZF}$, then we can run this argument without using reflection, of course. We can even pick $M$ to be an $L_\alpha$. But the point of picking $T$ finite is so that we can formalize this: If it were the case that Shoenfield's absoluteness result goes through without the requirement on $\omega_1$, then this would be provable in $\mathsf{ZF}$, and the proof of course only uses a finite set of axioms, and would apply to finite fragments of set theory, as long as they are strong enough. We can then let $T$ be so strong that, in particular, it contains all these axioms. This, naturally, leads to a contradiction, and the conclusion is that Shoenfield's theorem indeed needs the uncountability assumption.