If the distance between two consecutive terms in a sequence with bounded partial sum goes to 0, can we say the sequence converges to 0?

Here is a weird proof.

We are told that $|x_n|\leq c$ for some constant $c$. Suppose there is an $\epsilon>0$ such that $|x_n-x_{n+1}| \geq \epsilon$ for infinitely many $n$ (we reach a contradiction). Let $\{n[k]\}_{k=1}^{\infty}$ be a subsequence of indices for which $|x_{n[k]}-x_{n[k]+1}|\geq \epsilon$ for all $k \in \{1, 2, 3,…\}$. Fix $r\geq 10$ as a positive integer such that $(r-1)\epsilon >2c$ and consider the $r$-dimensional sequence $$\{(x_{n[k]}, x_{n[k]+1}, …, x_{n[k]+r-1})\}_{k=1}^{\infty}$$ This is bounded in $\mathbb{R}^r$ so the Bolzano-Wierstrass Theorem says there is a convergent subsequence defined by indices $n[k_m]$ such that $$ (x_{n[k_m]}, x_{n[k_m]+1}, …, x_{n[k_m]+r-1})\rightarrow (y_1, y_2, …, y_{r})$$ for some $(y_1, ..., y_r) \in \mathbb{R}^r$. Also, we must have $|y_1-y_r|\leq 2c$. On the other hand we know $x_n-2x_{n+1}+x_{n+2}\rightarrow 0$ so we must have $$ y_i - 2y_{i+1} + y_{i+2} = 0 \quad \forall i \in \{1, …, r-2\}$$ It follows that $y_i$ has the form: $$ \boxed{y_i = A + Bi \quad \forall i \in \{1, 2, 3, ..., r\}}$$ for some constants $A, B$. On the other hand we have $$|y_1-y_2|\geq \epsilon$$ and so $$ |\underbrace{(A+B)}_{y_1} - \underbrace{(A+2B)}_{y_2}|\geq \epsilon \implies |B|\geq \epsilon $$ But that means $$ |y_1-y_r| = |\underbrace{(A+B)}_{y_1}-\underbrace{(A+rB)}_{y_r}| =(r-1)|B|\geq (r-1)\epsilon >2c $$ which contradicts $|y_1-y_r|\leq 2c$. $\Box$