To simplify the notation, assume V=L. We have $|V_{\omega_{1}} |=\aleph_{\omega_{1}}$ and $|H(\aleph_{1})|=\aleph_{1}$, 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 enumerable 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.