Q:

Prove that $\lim_{n\to\infty}|f(x_n+\xi)-f(x_n)|=0$

Let $f:(0,+\infty)\to\mathbb{R}$ be continuous and bounded. Let $\xi>0$. Show that there is a sequence $(x_n)$ in $(0,+\infty)$ with $x_n\to\infty$ s.t.
$$\lim_{n\to\infty}|f(x_n+\xi)-f(x_n)|=0.$$
I tried to prove by contradiction. Assume this was false, then for any sequence $(x_n)$ with $x_n\to\infty$ and for any $n\in\mathbb{N}$, there exists $\varepsilon>0$ and $n_0\geqslant n$ s.t.
$$|f(x_{n_0}+\xi)-f(x_{n_0})|\geqslant\varepsilon$$
I want to conclude that in this case, $f$ cannot be bounded. I was stuck here. Am I on the right track? Can we prove this directly? I mean, without proof by contradiction.

A:

We have to prove that for all $x_0$, $\inf_{x\geqslant x_0}|f(x+\xi)-f(x)|=0$. If we manage to do that, then that $x_1$ such that $|f(x_1+\xi)-f(x)|\lt 1$,  $x_2\geqslant x_1+1$ such that $|f(x_2+\xi)-f(x_2)|\lt 1/2$ and more generally $x_{n+1}\geqslant x_n+n$ such that $|f(x_n+\xi)-f(x_n)|\leqslant n^{—1}$. 
This infimum exists and is finite since $f$ is bounded. Fix $x_0$ and let $\alpha:=\inf_{x\geqslant x_0}|f(x+\xi)-f(x)|$. Assume that $\alpha$ is positive. Define $g(x):=f(x+\xi)-f(x)$: $g$ is continuous doesn't vanish on $[x_0,\infty)$, hence, considering $-f$ instead of $f$, we can assume that $g(x)\geqslant \alpha\gt 0$ for all $x\geqslant x_0$. Using this with $x:=x_0+j\xi$, we get that 
$$f(x_0+n\xi)-f(x_0)\geqslant n\alpha.$$
Since $f$ is bounded, we get $\alpha\leqslant \frac{2\sup|f|}n$, hence $\alpha=0$, a contradiction. 

