Let $C_\alpha\subset\omega_1$ be a club set for every $\alpha<\omega_1$. Show that the diagonal intersection of all the $C_\alpha$'s, that is $\{\alpha<\omega_1:\forall\beta<\alpha,\alpha \in C_\beta\}$, contains a club set. I have a hint that says that assuming $f_\alpha:\omega_1\to\omega_1$ is continuous and order preserving with range $C_\alpha$ for every $\alpha$ (I showed that for every club set there exists a continuous and order preserving function whose range is the club set), define $g:(\omega_1)^2\to\omega_1$ as $g(\alpha,\beta)=f_\alpha(\beta).$
I'm not sure how to use the hint. Maybe take the restriction of g on the diagonal or something?
For each $\alpha < \omega_1$, define
$D_\alpha = \bigcap_{\xi\le \alpha} C_\xi$
$D_\alpha$ is closed and unbounded, since the intersection of less than $\omega_1$ closed unbounded subsets of $\omega_1$ is closed and unbounded.
It is easy to check that
$\Delta_{\alpha < \omega_1} C_\alpha = \Delta_{\alpha < \omega_1} D_\alpha$
Therefore, it suffices to work with the sequence $D_\alpha$.
First to show that $\Delta D_\alpha$ is closed. Let $\alpha$ be a limit point of $\Delta D_\alpha$. Suppose $\zeta < \alpha$. Let
$L = \{\mu \in \Delta \ D_\alpha : \zeta < \mu < \alpha\}$
By definition of diagonal intersection, $\mu \in D_\zeta$. Hence $L \subset D_\zeta$. $\sup L = \alpha$ since $\alpha$ is a limit point of $\Delta \ D_\alpha$. Hence $\alpha \in D_\zeta$ since $D_\zeta$ is closed. By definition of diagonal intersection, $\alpha \in \Delta D_\alpha$. $\Delta \ D_\alpha$ is closed.
Next to show that $\Delta \ D_\alpha$ is unbounded. Let $\beta < \omega_1$. Since $D_0$ is closed and unbounded, find $\gamma_0 > \beta$ such that $\gamma_0 \in D_0$. Recursively, define $\gamma_{n + 1} \in D_{\gamma_n}$ such that $\gamma_{n + 1} > \gamma_{n}$. Let $\gamma = \sup_{n \in \omega} \gamma_n$. Certainly $\gamma > \beta$. The claim is that $\gamma \in \Delta \ D_\alpha$. Let $\zeta < \gamma$. There exists some $k$ such that $\zeta < \gamma_k < \gamma$. $\gamma_{k + 1} \in D_{\gamma_k} \subseteq D_{\zeta}$ by definition of $D_\alpha$ and how the $\gamma_n$'s were constructed. Inductively, one can show that for all $n > k$, $\gamma_n \in D_\zeta$. Hence $\{\gamma_n : k < n \} \subseteq D_\zeta$. Hence $\gamma \in D_\zeta$ since $D_\zeta$ is closed. By definition of diagonal intersection, $\gamma \in \Delta \ D_\alpha$. $\gamma > \beta$, which was arbitrary. $\Delta \ D_\alpha$ is unbounded.
$\Delta D_\alpha = \Delta \ C_\alpha$ is club.