I would like to know why the following condition $\otimes$ is equivalent with the Continuum Hypotheses.
$\otimes$ There exists a sequence $\langle A_{\alpha} | \alpha < \omega_1\rangle$, such that for each $Z \subseteq \omega_1$ there exists two infinite ordinals $\alpha, \beta< \omega_1$ such that $Z \cap \beta = A_\alpha$. And, $A_\alpha \subseteq \alpha$ for each $\alpha < \omega_1$, of course.
(Taking $\beta:=\alpha$ yields the usual Jensen's $\large{\diamond}$, and this is logically stronger then CH.)
Only the nontrivial direction of the equivalence of CH with the above condition $\otimes$ is needed.
Both directions seem straightforward. If there is such a sequence, for any $Z\subseteq\omega$ there is an $\alpha$ such that $Z=A_\alpha$, since $Z=Z\cap\beta$ for any infinite $\beta$. This shows there is at most $\omega_1$ many subsets of $\omega$, that is, $\mathsf{CH}$ holds.
On the other hand, if $\mathsf{CH}$ holds, we have that the set of bounded subsets of $\omega_1$ has size $\aleph_1^{\aleph_0}=\aleph_1$, so we can list it as $(A_\alpha\mid\alpha<\omega_1)$ having each set appear unboundedly often, and ensuring that $A_\alpha\subseteq \alpha$ for all $\alpha$. Given any $Z\subseteq\omega_1$, for any $\beta$ we have that $Z\cap\beta$ is a bounded subset, so it appears as $A_\alpha$ for unboundedly many $\alpha$, so in particular for at least one infinite value of $\alpha$.