An Element in a C$^{*}$-algebra that is Almost a Projection is Close to a Projection Let $A$ be a C$^{*}$-algebra and let $\epsilon>0$ be given. I am trying to solve the following problem (Exercise 2.7. in Rordam's little blue book):

Show that there exists a $\delta>0$ with the following property: If $a\in A$ and $\|a-a^{*}\|\leq\delta$ and $\|a-a^{2}\|\leq \delta$, then there is a projection $p$ such that $\|a-p\|\leq \epsilon$. 

Following the hint, we assume that $\epsilon <1/2$ and set $b=\frac{a+a^{*}}{2}$. Then, it is not hard to see that $\sigma(b)\subset[-\epsilon,\epsilon]\cup[1-\epsilon,1+\epsilon]$ provided that $\|b-b^{2}\|\leq\epsilon-\epsilon^{2}$. The hint suggests to put $p:= f(b)$ for some continuous function $f$. Since we are assuming $\epsilon<1/2$, $\sigma(b)$ is disconnected, so I was thinking to make $f$ identically $0$ on $[-\epsilon,\epsilon]$ and identically $1$ on $[1-\epsilon,1+\epsilon]$ or vice-versa. Then, by the Spectral Mapping Theorem, $p$ would be a projection. However, I'm not sure how to conclude that $\|a-p\|\leq \epsilon$, or if this is even the correct choice of $p$. 
Also, presumably the fact that $\|b-b^{2}\|\leq\epsilon-\epsilon^{2}$, which we make use of follows from our choice of $\delta$ and the definition of $b$. However, in proving this, I came up with a $\delta$ in terms of $a$, but the way the question is asked suggests that the same $\delta$ should hold for all $a$ satisfying $\|a-a^{*}\|,\|a-a^{2}\|\leq\delta$. Is it possible to choose the $\delta$ independent of $a$?
Thank you very much!
 A: The inequality $\|a-p\|<\epsilon$ is simply $|x-f(x)|<\epsilon$ on $\sigma(b)$, i.e., $\|\text{id}-f\|<\epsilon$. 
As for the estimate: I assume that what you did is that if $\|a-a^*\|\leq\delta$ and $\|a-a^2\|\leq \delta$, then 
\begin{align}
\|b-b^2\|&=\left\|\frac{a+a^*}2-\left(\frac{a+a^*}2\right)^2\right\|\\ \ \\
&=\frac14\left\|{a-a^2}+\left({a-a^2} \right)^*+(a-aa^*)+(a^*-a^*a) \right\|\\ \ \\
&=\frac14\left\|{a-a^2}+\left({a-a^2} \right)^*+(a-a^2)+a(a-a^*)+(a-a^2)^*+a^*(a^*-a) \right\|\\ \ \\
&\leq \|a-a^2\|+\frac12\,\|a\|\,\|a-a^*\|\leq  \delta\,\left(1+\frac{\|a\|}2\right).
\end{align}
Now, since $b=b^*$, by computations you already made, $$\|b\|\leq\frac12+\sqrt{\frac14+\delta\,\left(1+\frac{\|a\|}2\right)}.$$ As $b$ is the real part of $a$, and $$\|\text{Im}(a)\|=\frac12\,\|a-a^*\|\leq\delta,$$
we have
$$
\|a\|\leq\|b\|+\frac\delta2\leq\frac{\delta+1}2+\sqrt{\frac14+\delta\,\left(1+\frac{\|a\|}2\right)}.
$$
We can rewrite this as 
$$
\|a\|^2-\left(\frac\delta2+1\right)\|a\|-\frac14-\delta+\left(\frac{\delta+1}2\right)^2\leq0.
$$
Then, analyzing the quadratic (i.e., trusting Wolfram Alpha),
$$
\|a\|\leq\frac14\,\left(2+3\delta+\sqrt{4+20\delta+5\delta^2}\right).
$$
