Exercise: Let $A$ be a nuclear $C^*$-algebra. Show that $A$ has Lance's WEP-property, i.e. show that there exists a ucp map $\Phi: B(H_u) \to A^{**}$ such that $\Phi(a)= a$ where $A\subseteq A^{**}\subseteq B(H_u)$ is the universal representation.
Attempt: If $A$ is unital, I can prove this. Indeed, choose ucp maps $$\varphi_k: A\to M_k(\mathbb{C}), \quad \psi_k: M_k(\mathbb{C})\to A$$ such that $\|\psi_k(\varphi_k(a)) -a\|\to 0$ for all $a\in A$. By Arveson's extension theorem, we may find ucp extensions $\widetilde{\varphi_k}: B(H_u)\to M_k(\mathbb{C})$. By compactness of the space of ucp maps $B(H_u)\to A^{**}$, the net $$\{\psi_k\circ \widetilde{\varphi_k}: B(H_u)\to A^{**}\}\subseteq UCP(B(H_u), A^{**})$$ admits a convergent subnet in the point-weak topology, say with limit $\Phi: B(H_u)\to A^{**}$. This map witnesses Lance's WEP property.
However, if $A$ is non-unital, then the above approach only gives a ccp map $\Phi: B(H_u)\to A^{**}$, and this map has to be unital. Is there a way to fix this? It is sufficient to show that if $A$ is non-unital, then $A$ has the WEP iff its unitisation $\widetilde{A}$ has the WEP. I can see that if $A$ has the WEP, then $\widetilde{A}$ has the WEP, but the converse is not clear to me.
For any $a\in A^+$ with $\|a\|\leq1$ we have, since $0\leq a\leq 1$ and $\Phi$ is positive, $$ a=\Phi(a)\leq\Phi(1). $$ By Kaplansky the selfadjoint part of the unit ball of $A^{**}$ is the sot-closure of the selfadjoint part of the unit ball of $A$, and thus $\Phi(1)\geq b$ for all $b\in (A^{**})_1$. In particular $\Phi(1)\geq1$. As $\Phi(1)$ is positive with norm at most $1$, we also have $\Phi(1)\leq1$, so $\Phi(1)=1$.