Continuity and diverging sequences

Hint: Suppose for some $S$, there does not exist a sequence $(x_n)$ such that $\lim x_n = \infty$ and $\lim f(x_n+S) - f(x_n) = 0$, and let $g(x) = f(x + S) - f(x)$. Then, we must have that $g$ is bounded away from $0$ for $x$ sufficiently large, say, greater than $x_0$. (Why?) What does this tell us about the sequence $(f(x_0), f(x_0+S), f(x_0 + 2S), \cdots)$?

You can easily draw functions which are bounded, not periodic and that have no limit at infinity. And even if your function were periodic, the exercise would not be simpler if $S$ has nothing to do with the period of your function. So your division of the problem is not very useful.

The real number $S$ that is given can be seen as a jump. You have to compare the values of your function evaluated at two points separated by this jump. And you want to show that it can be very small.

Could it be arbitrary large? Certainly not, since your function is bounded. But we want to show that this difference $f(x+S)-f(x)$ can be arbitrary small... Assume by contradiction, that $f(x+S)-f(x)$ is always greater than some number, say $1$, in absolute value.

Hence, each jump of $S$ implies a jump of $1$ in the values of the function. This seems to imply that the function is unbounded, right? No, there is still a problem, because one could maybe have jump greater than $1$ and then jumps less than $-1$, in some alternating way...

If this happens, you can use that your function is continuous. I let you deal with the details...