Diamond implies CH

The tail $C=\{\alpha <{\omega }_1:\alpha >\omega \}$ is club: it is unbounded, and any nonzero limit point of it is greater than $\omega$ and still belongs to it. For $B\subseteq \omega$, apply F1 to $B$ and $C$. At the resulting $\alpha >\omega$ one has $B\cap \alpha =B$, so $A_{\alpha }=B$. Define $j(B)$ to be the least such $\alpha >\omega$. This minimum exists in the nonempty set of eligible ordinals. If $j(B)=j(D)=\alpha$, then $B=A_{\alpha }=D$, proving that $j:\mathcal{P}(\omega )\to {\omega }_1$ is injective. The argument includes $B=\mathrm{\varnothing }$ and $B=\omega$.

Step 1.1 gives $\mathrm{\mid }\mathcal{P}(\omega )\mathrm{\mid }\le {\mathrm{\aleph }}_1$. F3 with $A=\omega$ gives ${\mathrm{\aleph }}_0<\mathrm{\mid }\mathcal{P}(\omega )\mathrm{\mid }$, and by F4 and A1 any uncountable cardinal is at least the least uncountable cardinal ${\mathrm{\aleph }}_1$. The two inequalities give $\mathrm{\mid }\mathcal{P}(\omega )\mathrm{\mid }={\mathrm{\aleph }}_1$, which is $2^{{\mathrm{\aleph }}_0}={\mathrm{\aleph }}_1$.