Suppose $0 < x < 1 $ and let $n > 0 $ such that $\frac{1}{n+1} \leq x < \frac{1}{n}$. If there exists a sequence $x_n \downarrow x$, then there exists $N$ such that $\frac{1}{n+1} \leq x_n < \frac{1}{n}$ for all $n \geq N$
MY try:
Let $\epsilon = 1/n$ so we have and $N$ such that for all $n >N$,
$$ x - 1/n < x_n < x + 1/n $$
How can I bound this inequality to obtain the result?