All rational numbers between 0 and 1 inclusive will eventually yield some $x_k = 0$ . To begin, note that by definition, all rational numbers can be written as a quotient of coprime integers. Let $x_0 = \frac{m}{n}$ , where $m,n$ are coprime positive integers. Since $0<x_0<1$ , $0<m<n$ . Now $$x_1 = \left\{\frac{p_1}{\frac{m}{n}}\right\}=\left\{\frac{np_1}{m}\right\}.$$ From this, we can see that applying the iterative process will decrease the value of the denominator, since $m<n$ . Moreover, the numerator is always smaller than the denominator, thanks to the fractional part operator. So we have a strictly decreasing denominator that bounds the numerator. Thus, the numerator will eventually become 0.
On the other hand, if $x_0$ is irrational, then a simple induction will show that $x_k$ will always be irrational. Indeed, the base case has been established, and, if $x_k$ is irrational, then $\dfrac{p_{k+1}}{x_k}$ must be too, and likewise for its fractional portion, which differs from it by an integer. Hence $x_{k+1}$ is irrational, completing the proof.