Proof of Glimm's Lemma I am trying to understand the proof of Glimm's lemma from Brown & Ozawa p.8:

Theorem 1.4.11: Let $A\subseteq B(H)$ be a separable $C^*$-algebra
  containing no non-zero compact operators. If $\varphi$ is a state on
  $A$ , then there exist orthonormal vectors ($\xi_n$) such that
  $\langle a\xi_n,\xi_n\rangle\to \varphi(a)$ for all $a\in A$

I won't write here the proof, it can be found in the book, but here are my questions:
$1.$ "By excision, for each k, there exists a norm-one positive element..." Why norm-one? in Theorem $1.4.10$ (Excision) we don't get norm-one elements. And if their norm is very small, I think I can't just divide by the norm, right? Presumably it is connected to the fact that $F$ is a finite set?
$2.$ "...notice that $P_{K_0}^{\perp}e_1P_{K_0}^{\perp}-e_1$ is ac compact operator, hence $||P_{K_0}^{\perp}e_1P_{K_0}^{\perp}||=1$.
I agree it is compact, and my first intuition was that $A$ has no compact operators, so it must be zero and that's all. But maybe this operators is not in $A$. So, why the norm must be equal to $1$?
$3.$ "Let $\zeta_1\in  K_0^{\perp}$ be a unit vector s.t. $||e_1\zeta_1-\zeta_1||<\epsilon$. Seems like a direct conclusion from the last paragraph, why is that true?
 A: 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 $$\frac{\delta^{1/2}+1-(1-\delta)^{1/2}}{(1-\delta)^{1/2}}<\varepsilon.$$
A: More conceptual for point 3!

Observation:
  Suppose $v:X\to Y$ is an isometry (e.g. of Hilbert spaces). Then
  $$\|vT\|=\|T\|\text{ and so also }\|Tv^*\|=\|T\|$$
  for every (adjointable) operator $T:X'\to X$.

Consider the embedding and projection
$$v:V^\perp\to H\text{ and }
v^*:H\to V^\perp.$$
In what follows, think of them really as
$$
v=\begin{pmatrix}0\\1\end{pmatrix}\text{ and }v^*=\begin{pmatrix}0&1\end{pmatrix}.
$$
Using the observation, we have from point 2
$$
1=\|QeQ\|=\|vv^*evv^*\|=\left\|v^*ev\right\|
$$
where $Q=
\left(\begin{smallmatrix}0\\1\end{smallmatrix}\right)
\left(\begin{smallmatrix}0&1\end{smallmatrix}\right)
= vv^*$.
Denote the numerical range by
$$
\mathcal{W}(T):=\{\langle Tx,x\rangle:\|x\|=1\}.
$$
It holds for normal and so also positive operators $T\geq0$,
$$
\|T\|=r(T)
\text{ and }
\overline{\mathcal{W}(T)}=\operatorname{conv}\sigma(T).
$$
Since $\left\|v^*ev\right\|=1$ we may thus find $x_n\in H$ with $\|x_n\|=1$ such that
$$
\langle vev^* x_n,x_n\rangle\to1
$$
See where we are heading to? Let's evaluate.
Put equivalently, we may find $x\in H$ with $\|x\|^2=1$ such that
$$
1-\langle v^*ev x,x\rangle<\epsilon.
$$
Note also that since $0\leq e\leq1$,
$$
(1-e)^2\leq(1-e).
$$
With this at hand, we obtain
\begin{align}
\|(1-e)vx\|^2&=\langle(1-e)\ldots,(1-e)\ldots\rangle \\
&=\langle(1-e)^2\ldots,\ldots\rangle \\
&\leq\langle(1-e)\ldots,\ldots\rangle \\
&=\langle vx,vx\rangle -\langle v^*evx,x\rangle
=1-\langle v^*evx,x\rangle<\epsilon
\end{align}
and where $\langle vx,vx\rangle=1$ since $v^*v=1$.
