To simplify the notation, assume $V=L$. We have $\lvert V_{\omega_{1}} \rvert=\aleph_{\omega_{1}}$ and $\lvert H(\aleph_{1})\rvert=\aleph_{1}$, so in particular $V_{\omega_{1}} \models \exists x \forall \alpha\; x \not\in L(\alpha) $ since $L(\omega_{1})=H(\aleph_{1})$.

Using the Löwenheim–Skolem theorem we have a transitive countable set $M\prec V_{\omega_{1}}$. In particular, $M\in H(\aleph_{1})$. We have $H(\aleph_{1})\prec_{1}V_{\omega_{1}}$, so $M\prec_{1}H(\aleph_{1})$. By the condensation lemma $M=L(\alpha)$. $\alpha$ must be a limit ordinal, but $L(\beta)\in L(\beta+1)$, so $L(\alpha)\models \forall x\exists \beta \; x\in L(\beta)$ for all ordinal limit $\alpha$. This is a contradiction with $M\prec V_{\omega_{1}}$.

Where is the mistake? I didn't find it.

Edit: I think the Mostowski collapse lemma is not the core of problem. We have that exist a $\alpha$ limit with $V_{\omega_{1}}\in L(\alpha)$, and we have $V_{\omega_{1}}\prec_{1}L(\alpha)$, but as $V_{\omega_{1}}$ is transitive his transitive collapse is himself, but is not possible.

The heart of problem is the relation $V_{\omega_{1}}\prec_{1} L(\alpha)$, but why is it wrong?

Edit: thanks the comentaries the solution is, despite $V_{\omega_{1}}$ and $L(\alpha)$ satisfies the same $\Sigma_{1}$ sentences $\varphi$, we need to consider all formulas $\varphi(x)$ that are $\Sigma_{1}$

4more comments