Let $x_n=\frac{k}{2^{s+1}}$ for $n=2^{s+1}+k$, $k<2^s$.
In the other words, you define the sequence separately for $\{1\}$, $\{2,3\}$, $\{4,5,6,7\}$, $\ldots$, $\{2^s,2^s+1,\ldots,2^{s+1}-1\}$.
The step between the neighbors is always $\frac1{2^{s+1}}\le \frac1{2^s+k} = \frac1n$, only at the end of each block you jump below.
Now it remains to show that this sequence is not convergent. Can you show this?
A different example could be: $x_n=\sin t_n$ where $t_n=\sum_{k=1}^n \frac1k$ is the $n$-th partial sum of the harmonic series.
You have $|x_{n+1}-x_n| = |\sin t_{n+1}-\sin t_n| \le |t_{n+1}-t_n| = \frac1{n+1}\le \frac1n.$
(We have used $|\sin(a-b)| \le |a-b|$, see e.g. here.)
The inequality $|x_{n+1}-x_n|\le\frac1n$ clearly implies $x_{n+1}\le x_n+\frac1n$.