V equals L implies diamond

Define $(S_{\alpha }{)}_{\alpha <{\omega }_1}$ by F6. Having defined the earlier guesses, call $(B,D)$ bad at $\alpha$ when $B\subseteq \alpha$, $D\subseteq \alpha$ is club in $\alpha$, and $S_{\xi }\mathrm{\ne }B\cap \xi$ for every $\xi \in D$. If a bad pair exists, take the $\chi$-least ordered pair and set $S_{\alpha }=B$; otherwise set $S_{\alpha }=\mathrm{\varnothing }$. F2 makes the choice unique and F1 makes this one fixed first-order recursion; $S_0=\mathrm{\varnothing }$, and always $S_{\alpha }\subseteq \alpha$.

Assume for contradiction that this sequence is not diamond. By F7 there are $A\subseteq {\omega }_1$ and a club $C\subseteq {\omega }_1$ such that $S_{\alpha }\mathrm{\ne }A\cap \alpha$ for every $\alpha \in C$. Among all such global failure pairs choose the $\chi$-least $(A,C)$, possible because $V=L$ and F2 well-orders $L$.

Choose a nonzero limit $\theta$ such that $L_{\theta }$ contains $S,A,C,{\omega }_1$. Because $L_{\theta }$ is a ${<}_L$-initial segment and the badness predicate has only bounded quantifiers once these parameters are fixed, it sees that $(A,C)$ is the least failure pair. Let $X$ be the canonical hull of this finite seed. F3 gives $X\prec L_{\theta }$ and makes $X$ countably infinite. Put $\delta =X\cap {\omega }_1$. Elementarity makes $X\cap {\omega }_1$ an ordinal, hence transitive, and countability gives $\delta <{\omega }_1$. For every $\xi <\delta$, elementarity applied to the unbounded set $C$ produces $c\in C\cap X$ above $\xi$; hence $C\cap \delta$ is unbounded in $\delta$. It follows that $\delta$ is a nonzero countable limit, and closure of $C$ gives $\delta \in C$.

Collapse $X$ by $\pi$ to $M=L_{\gamma }$ using F4. By F5, $\pi ({\omega }_1)=\delta$. Since every $\xi <\delta$ lies in $X$, evaluation of the function $S\in X$ puts $S_{\xi }\in X$; as $S_{\xi }\subseteq \xi$, F5 fixes it. The collapse equations therefore give $\pi (S)=S\upharpoonright \delta$, $\pi (A)=A\cap \delta$, and $\pi (C)=C\cap \delta$.

By elementarity and isomorphism, $M$ regards $(A\cap \delta ,C\cap \delta )$ as its $\chi$-least failure pair for the sequence $S\upharpoonright \delta$ on its first uncountable ordinal $\delta$. The predicates “subset of $\delta$,” “club in $\delta$,” and “fails at every member” are bounded here and are absolute between the transitive $M$ and the universe for these fixed parameters. F1 says that $M$ and the universe use the same $\chi$-order on $M$, and that $M=L_{\gamma }$ is an initial segment of that order. Therefore no ambient bad pair at $\delta$ can precede $(A\cap \delta ,C\cap \delta )$: any preceding pair would belong to $M$ and contradict internal leastness. Thus step 1.1 sets $S_{\delta }=A\cap \delta$.

But step 3.1 gives $\delta \in C$, while step 2.1 says $S_{\delta }\mathrm{\ne }A\cap \delta$ at every member of $C$. This contradicts step 5.1. Hence the sequence is diamond, and its correct-guess set meets every club for every target subset of ${\omega }_1$.