= Solution
One can take $\boxed{\alpha=\omega+1}$. Finite induction gives $L_n=V_n$ for every finite $n$: every subset of a finite level is definable using its finitely many elements as parameters. Hence $L_\omega=V_\omega$, the set of <hereditarily finite sets>.
Now reason inside the <constructible universe> $L$, a model of <ZFC>. The set $L_\omega$ is countable there, and there are countably many <first-order formulas> and finite parameter tuples. Thus $L_{\omega+1}=\operatorname{Def}(L_\omega)$ is countable in $L$. But the <constructible power set> $\mathcal P^L(\omega)$ is uncountable in $L$ by the <Cantor theorem>. Choose
$$
r\in\mathcal P^L(\omega)\setminus L_{\omega+1}.
$$
This $r$ is constructible and a subset of $\omega$, so $r\subseteq V_\omega$ and $r\in V_{\omega+1}$. Therefore
$$
\boxed{r\in L\cap V_{\omega+1}\setminus L_{\omega+1}}.
$$
All comparisons were made inside $L$; this avoids assuming that the constructible reals are uncountable in the ambient universe. The witness remains valid externally because the two structures use the same $r$, the same $V_\omega$, and the same constructible level. Thus <constructible sets of low rank can appear at later stages>.
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