Is the following argument correct?
Proposition. Suppose $\{x_n\}$ is a sequence and suppose for some $x\in\mathbf{R}$, the limit $$L:=\lim_{n\to\infty}\frac{|x_{n+1}-x|}{|x_n-x|}$$ exists and $L<1$. Show that $\{x_n\}$ converges to $x$.
Proof. We choose an $r$ such that $L<r<1$, then in particular for $\epsilon = r-L$, there exists an $M\in\mathbf{N}$, such that $\left|\frac{|x_{n+1}-x|}{|x_n-x|}-L\right|<r-L,\forall n\ge M$ and by extension $\frac{|x_{n+1}-x|}{|x_n-x|}-L<r-L,\forall n\ge M$. We may therefore surmise that $$\frac{|x_{n+1}-x|}{|x_n-x|}<r,\forall n\ge M$$ Now given an arbitrary $n>M$, the above proposition implies the following \begin{align*} |x_n-x| &= |x_M-x|\frac{|x_{M+1}-x|}{|x_{M}-x|}\frac{|x_{M+2}-x|}{|x_{M+1}-x|}\cdots\frac{|x_n-x|}{|x_{n-1}-x|}\\ &<|x_M-x|rr\cdots r = |x_M-x|r^{n-M} = (|x_M-x|r^{-M})r^{n}. \end{align*} consequently the sequence $\{x_{n+M}\}_{n=1}^{\infty}$ is such that we have $$|x_{n+M}-x|< (|x_M-x|r^{-M})r^{n},\forall n\in N$$ Now since $r<1$, $\lim_{n\to\infty}r^n = 0$ and by extension $\lim_{n\to\infty}(|x_M-x|r^{-M})r^{n}= 0$. Finally appealing to proposition $\textbf{2.2.4}$ implies that $\{x_{n+M}\}_{n=1}^{\infty}$ and by extension $\{x_n\}$ converge to $x$.
$\blacksquare$
Note: Proposition $\textbf{2.2.4}$ is the result that if $x\in\mathbf{R}$ and we have a sequence $\{a_n\}$ such that $\lim_{n\to\infty}a_n = 0$,and we have a sequence $|x_n-x|<a_n,\forall n\in\mathbf{N}$, then $\lim_{n\to\infty}x_n = x$.
Yes. Your proof is right, but you also can use another simpler proof. Define $y_n=|x_n-x|$. Then according to Ratio test for $y_n$ we know that $\sum y_n$ exists and is bounded since $$0<\lim_{n\to \infty}{y_{n+1}\over y_n}=L<1$$because $\sum y_n$ exists and is bounded it must be necessarily that $y_n\to 0$ or equivalently $x_n\to x$