$y[n] = x[n-1]$ is an all pass filter I'd like to know if I understand correctly why $y[n] = x[n-1]$ is an all pass filter.

By definition of the filter, we get the output the same of the input only delayed by one unit of time. Therefore, all frequencies that entered must pass (eventually). 

Therefore, it's an all-pass filter.

Is that a correct reasoning? 

Thanks
Your notation is a bit odd, as you are mixing the time-domain with the $z$-domain in the equation you wrote. If you refer to the system

$$y[n]=x[n-1] \quad \mathrm{or} \quad Y(z)=z^{-1}X(z)$$

then you are correct. To put it mathematically, the transfer function of the filter would be:

$$H(z)=\frac{Y(z)}{X(z)}=\frac{1}{z}$$

The frequency response is then

$$H(z) \Bigg|_{z=e^{j\omega}} =\frac{1}{e^{j\omega}}$$

You can see that the magnitude of $H(e^{j\omega})$ is $1$ for all frequencies. Therefore, it is an all-pass system.