In their Berkeley Lectures, to motivate the introduction of Diamonds, Scholze and Weinstein discuss what should be the definition of $\operatorname{Spa}\mathbf{Z}_p\times \operatorname{Spa}\mathbf{Z}_p$. The analogy is made with the "equal characteristic" setting, where the counterpart of $\mathbf{Z}_p$ is $\mathbf{F}[\![t]\!]$ for a certain finite field $\mathbf{F}$ and an indeterminate $t$. Then $\operatorname{Spa} \mathbf{F}[\![t]\!]\times_{\operatorname{Spa} \mathbf{F}}\operatorname{Spa} \mathbf{F}[\![t]\!]$ equals $\operatorname{Spa} \mathbf{F}[\![t,u]\!]$ where we named $u$ the right-hand indeterminate.
In that regard, why isn't $\operatorname{Spa}\mathbf{Z}_p\times \operatorname{Spa}\mathbf{Z}_p$ something as simple as $\operatorname{Spa} W_p(\mathbf{Z}_p)$ ? (Sorry if this is a naive question).
EDIT: I shall add more precisions to my question. As quoted from the Berkeley Lectures, $\operatorname{Spa}\mathbf{Z}_p\times \operatorname{Spa}\mathbf{Z}_p$ should contain $\operatorname{Spa}\mathbf{Q}_p\times \operatorname{Spa}\mathbf{Q}_p$ as a dense subset, so Scholze and Weinstein first construct the latter (on page 5). By analogy with the function fields side where $\operatorname{Spa} \mathbf{F}(\!(t)\!)\times_{\operatorname{Spa} \mathbf{F}}\operatorname{Spa} \mathbf{F}(\!(t)\!)$ is the punctured open unit disc over $\mathbf{F}(\!(t)\!)$, they consider the punctured open unit disc $D_{\mathbf{Q}_p}^*$ over $\mathbf{Q}_p$. Next comes two steps that I do not understand:
They consider the limit $\tilde{D}_{\mathbf{Q}_p}=\varprojlim D_{\mathbf{Q}_p}$ where the transition maps are $x\mapsto (x+1)^p-1$.
They consider the quotient $\tilde{D}_{\mathbf{Q}_p}^*/\mathbf{Z}_p^{\times}$.
In definition 1.2.1, they define $\operatorname{Spa}\mathbf{Q}_p\times \operatorname{Spa}\mathbf{Q}_p$ as $\tilde{D}_{\mathbf{Q}_p}^*/\mathbf{Z}_p^{\times}$. But why to not stop at $D_{\mathbf{Q}_p}^*$?