Technical lemma about state space of a $C^*$-algebra. Consider the following proof from the book "$C^*$-algebras and finite-dimensional approximations":

Why does this proof work in the non-unital case? (see the last line). Maybe we have
$$\|a + \lambda \| = \sup_{\varphi \in \mathcal{S}} \{|\varphi(a, \lambda)|\}$$
where the states $\varphi: A  \to \mathbb{C}$ have been extended (uniquely) to states on the unitization?
 A: The states of $A$ extend uniquely (as states) to the unitization. The extension is $\tilde\varphi(a+\lambda)=\varphi(a)+\lambda$. It easy to check that such extension is a state.
And you have an extension $\psi$ of $\varphi$ that is a state, then $\psi(1)=1$ (a positive functional always satisfies $\psi(1)=\|\psi\|$). That gives you
$$
\psi(a+\lambda)=\psi(a)+\psi(\lambda)=\varphi(a)+\lambda. 
$$
The only state in the unitization $\tilde A$ of $A$ that does not come from a state in $A$ is the state $\alpha(a+\lambda)=\lambda$. As this state is zero in all elements of $A$, it not in the weak-$*$ closed convex hull of the other states, so it doesn't affect the definition of $S$ nor the proof.
I don't see a way to go about the proof in an obvious way as Nate and Taka make it sound, but here is my take. Write $n=\|a\|$. Then, with $\tilde\varphi$ the unique extension of $\varphi$ to the unitization as a state,
\begin{align}
\|a+n\|&=\sup_{\|b\|\leq1}\|ab+nb\|
=\sup_{\|b\|\leq1}\sup_{\varphi\in S}|\varphi(ab+nb)|\\[0.3cm]
&=\sup_{\|b\|\leq1}\sup_{\varphi\in S}|\tilde\varphi((a+n)b)|\\[0.3cm]
&\leq\sup_{\|b\|\leq1}\sup_{\varphi\in S}\tilde\varphi(b^*b)^{1/2}\tilde\varphi((a+n)^2)^{1/2}\\[0.3cm]
&\leq\sup_{\varphi\in S}\tilde\varphi(\|a+n\|\,(a+n))^{1/2}\\[0.3cm]
&=\|a+n\|^{1/2}\,\sup_{\varphi\in S}\tilde\varphi(a+n)^{1/2}.
\end{align}
Cancelling and squaring we obtain
$$
\|a+n\|\leq\sup_{\varphi\in S}\tilde\varphi(a+n),
$$
which is the non-obvious inequality that is needed.
