Please help prove that $(x_n)$ is a Cauchy sequence if $|x_{n+1} - x_n| \leq Cr^n$

Let $\epsilon>0$ be given. Then we can find $N$ such that $r^N\frac C{\color{red}{1-r}}<\frac\epsilon2$ because $r^n\color{red}\to0$ as $n\to \infty$. Then for any $n>N$ we have $$\begin{align} |x_n-x_N|&\le|x_{N+1}-x_N|+\ldots+ |x_n-x_{n-1}|\\&\le Cr^N+Cr^{N+1}+\ldots +Cr^{n-1} \\&=Cr^N\cdot(1+r+\ldots + r^{n-N-1}) \\&\color{red}<Cr^N\sum_{k=0}^\infty r^k \color{red}= Cr^N\cdot\frac1{1-r}<\frac\epsilon2,\end{align}$$ hence for $n,m>N$ $$ |x_n-x_m|\le|x_n-x_N|+|x_m-x_N|<\frac\epsilon2+\frac\epsilon2=\epsilon.$$

Remark: I've marked all places in red where we made use of $0<r<1$.

Hint: $|x_n-x_m|\le |x_n- x_{n-1}|+|x_{n-1}-x_{n-2}|+\dots+|x_{m+1}-x_{m}|$.