I am referring to the proof of (4) implies (1) in Theorem 3.16 of Woodin's *The Axiom of Determinacy, Forcing Axioms, and the Nonstationary Ideal*. His proof leverages on the fact that if the sharp of every real exists, then $\delta^1_2 = u_2$, where $u_2$ is the second uniform indiscernible (the least ordinal above $\omega_1$ which is an $x$-indiscernible for every real $x$).

This is roughly how it goes:

- Fix any ordinal $\alpha$ strictly between $\omega_1$ and $\omega_2$, and a well-ordering $<_{\alpha}$ of $\omega_1$ of ordertype $\alpha$.
- Find a club $C$ of $\omega_1$ such that for every $\gamma \in C$, the rank of $<_{\alpha}$ restricted to $\gamma$ is less than the least ordinal in C greater than $\gamma$.
- By hypothesis, there is a $D \subset C$, $D$ a club of $\omega_1$, such that $D$ is constructible from a real $z$. We can assume $D$ is definable from $z$ and $\omega_1$ in $L[z]$ since $z^\sharp$ exists.
- By reflecting the definition of $D$ downwards, we have that for every $\gamma \in D$, $rank(<_{\alpha} \restriction \gamma) < rank(\mathcal{M}(z^{\sharp}, \gamma + 1))$, where $\mathcal{M}(z^{\sharp}, \gamma + 1)$ stands for (quoting Woodin) "the $\gamma$[$+1$] model of $z^{\sharp}$". I am not exactly sure what that quote means, so I assumed $rank(\mathcal{M}(z^{\sharp}, \gamma + 1))$ is the least $z$-indiscernible above $\gamma$, which works for the arguments hitherto. Did I make a mistake in my interpretation here?
- Now, following the previous inequality of ranks, he immediately concluded that $\alpha < rank(\mathcal{M}(z^{\sharp}, \omega_1 + 1))$, which finishes the proof.

It is at the last step (the final bullet point) that I got lost. How does $\alpha < rank(\mathcal{M}(z^{\sharp}, \omega_1 + 1))$ follow from the previous steps?

EDIT: Alternatively, I would appreciate it if anyone can point me to another proof of the statement "if the sharp of every real exists and every club contains a club constructible from a real, then $\delta^1_2 = \omega_2$"