In the proof of $(a) \Rightarrow(b)$, we have $\pi_{\omega}(f(\phi(1))=0$. It contradicts the fact that the spectrum of $\pi_{\omega}((\phi(1))$ is $[0, 1]$. I don't understand this statement. I know the fact the spectrum of $\pi_{\omega}(f(\phi(1))$ is 0, but $f$ is not any element in $C_0((0,1])$, $f$ may not be the $id_{(0,1]}$.
Since $\sigma(\pi_\omega(\phi(1)))=[0,1]$, the spectral mapping theorem tells us that $$\{0\}=\sigma(\pi_\omega(f(\phi(1))))=f(\sigma(\pi_\omega(\phi(1))))=f([0,1]).$$ (Here, $\sigma(x)$ denotes the spectrum of the element $x$.) The only $f\in C_0((0,1])$ mapping the whole interval $[0,1]$ to $\{0\}$ is the constant zero function. But by hypothesis, $f$ is assumed to be non-zero.