Proof of Glimm's Lemma

It is fair to say that Nate and Taka omitted "a few details".

In Theorem 1.4.10, you have that $0\leq e_i\leq1$, so $\|e_i\|\leq1$. And you also have $\varphi(e_i)=1$. Since $\varphi$ is a state, $$ 1=\varphi(e_i)\leq\|\varphi\|\,\|e_i\|=\|e_i\|. $$ So $\|e_i\|=1$.

Since $A$ contains no compact operators, the restriction of the quotient map $\pi:B (H )\to B(H)/K(H)$ to $A $ is one-to-one; being a $*$-homomorphism, it is isometric. Then $1=\|e_1\|=\|\pi(e_1)\|$. Also, as $P_{{\mathcal K}_0}^\perp e_1 P_{{\mathcal K}_0}^\perp-e_1$ is compact, it is in the kernel of $\pi $; so we have $$ \pi(P_{{\mathcal K}_0}^\perp e_1 P_{{\mathcal K}_0}^\perp)=\pi(e_1). $$ It follows that $$ 1\geq\|P_{{\mathcal K}_0}^\perp e_1 P_{{\mathcal K}_0}^\perp\|\geq\|\pi(P_{{\mathcal K}_0}^\perp e_1 P_{{\mathcal K}_0}^\perp)\|=\|\pi(e_1)\|=1. $$

For notation simplicity, write $Q=P_{\mathcal K_0}^\perp$. Fix $\delta>0$. Using that $1\in\sigma( Q e_1 Q)$, let $\zeta_0\in H$ be a unit vector such that $\|Qe_1Q\zeta_0-\zeta_0\|<1-(1-\delta)^{1/2}$. In particular $$1-\|Qe_1Q\zeta_0\|=\|\zeta_0\|-\|Qe_1Q\zeta_0\|\leq\|\zeta_0-Qe_1 Q\zeta_0\|\leq1-(1-\delta)^{1/2},$$ so $$ (1-\delta)^{1/2}\leq\|Qe_1Q\zeta_0\|\leq1. $$ Then $$\tag{1} 1-\delta\leq\|Qe_1Q\zeta_0\|^2\leq1. $$ Also, $$\tag{2} 1-\delta\leq\|Qe_1Q\zeta_0\|^2\leq\|e_1Q\zeta_0\|^2\leq1. $$

We have $$ \|Qe_1Q\zeta_0\|^2+\|Q^\perp e_1Q\zeta_0\|^2=\|Qe_1Q\zeta_0+Q^\perp e_1Q\zeta_0\|^2=\|e_1Q\zeta_0\|^2, $$ so from $(2)$ we get $$\tag{3} 1-\delta\leq \|Qe_1Q\zeta_0\|^2+\|Q^\perp e_1Q\zeta_0\|^2\leq1. $$ From $(2)$ and $(3)$, we get that $$ \|Q^\perp e_1Q\zeta_0\|^2<\delta. $$ We also have $$ \|Q^\perp\zeta_0\|=\|Q^\perp(\zeta_0-Qe_1Q\zeta_0)\|<1-(1-\delta)^{1/2}, $$ so $$\tag{4} \|Q\zeta_0\|=\|\zeta_0-Q^\perp\zeta_0\|\geq1-\|Q^\perp\zeta_0\| \geq1-(1-(1-\delta)^{1/2})=(1-\delta)^{1/2}. $$ Then \begin{align} \|e_1Q\zeta_0-Q\zeta_0\|&\leq\|Q^\perp e_1Q\zeta_0\|+\|Qe_1Q\zeta_0-Q\zeta_0\|\\ \ \\ &\leq\delta^{1/2}+\|Q(Qe_1Q\zeta_0-\zeta_0)\|\\ \ \\ &\leq\delta^{1/2}+\|Qe_1Q\zeta_0-\zeta_0\|\\ \ \\ &\leq\delta^{1/2}+1-(1-\delta)^{1/2}. \end{align} So letting $\zeta_1=Q\zeta_0/\|Q\zeta_0\|$, we have $$ \|e_1\zeta_1-\zeta_1\|=\frac{\|e_1Q\zeta_0-Q\zeta_0\|}{\|Q\zeta_0\|} \leq\frac{\delta^{1/2}+1-(1-\delta)^{1/2}}{(1-\delta)^{1/2}}, $$ where the estimate for the denominator comes from $(4)$. Now all that's left is to choose $\delta$ so that $$\frac{\delta^{1/2}+1-(1-\delta)^{1/2}}{(1-\delta)^{1/2}}<\varepsilon.$$