I'm currently reading the paper "The nuclear dimension of $C^*$-algebras" by Winter and Zacharias.
I'm trying to understand the proof of:

Proposition 3.2.: Let $A$ be a $C^*$-algebra with $dim_{nuc}A=n<\infty$. Then, there is a system $(F_{\lambda},\psi_{\lambda},\varphi_{\lambda})_{\lambda\in\Lambda}$ of piecewise contractive $n$-decomposable c.p. approximations such that the map: $\bar{\psi}:A\to \Pi_{\Lambda}F_{\lambda}/\oplus_{\Lambda}F_{\lambda}$ induced by the $\psi_{\lambda}$ has order zero.

Proof: Assume first that $A$ is separable. In this case, it will suffice to show the following: For any $\epsilon>0$ and any finite subset $\cal{F}\subseteq A$ of positive normalized elements, there is a piecewise contractive $n$-decomposable c.p. approximation $(F,\psi,\varphi)$ such that:

1. $||\varphi\psi(b)-b||<\epsilon^{1/16}$ for $b\in \cal{F}$; and
   2.$||\psi(c)\psi(c')||<\epsilon^{1/16}$ whenever $c,c' \in \cal{F}$ satisfy $||cc'||<\epsilon$.
   

My question is how do we use the separability assumption?
For example, if one needs to show that $dim_{nuc}A\leq n$ then it suffices to show it locally, as done above, without assuming $A$ is separable.
Therefore, I think that the second condition uses separability and am not sure why.

Moreover, why do we need to consider $||cc'||<\epsilon$ instead of $cc'=0$?

Thank you for any help!