Analysis Prelim Question, showing $f(ny)\to 0$, as $n\to \infty$ implies $f(y)\to 0$ as $y\to \infty$

Question

In preparing for the quals, I stumbled across this question:
Let $f$ be a continuous real-valued function on $[0,\infty)$ with $f(0) = 0$. Suppose that for each $y\in [0, 1]$ we have that $f(ny)\to 0$ as $n\to \infty$ through the integers. Prove that $f(x)\to 0$ as $x\to \infty$ through the reals.
So, In trying to prove it I knew that if it were not the case then for some $\epsilon > 0$ we have that there exist non-degenerate closed intervals, $A_n = [a_n, b_n]$ such that $a_n \rightarrow \infty$ and $f(x) > \epsilon$ for $x \in A_n.$
Perhaps then, I thought, an infinite number of $A_n$ non-trivially intersect, for a fixed $k$, $R_k = \{$Rationals with denominator $k\} $. If this were the case, then $f(n/k)$ would not got to 0 as $n \rightarrow \infty$.
Unfortunately~! My attention was then brought to the counterexample of $A_k = [k + 1/k, k + 2/k]$...
Darn.
So, I guess that I need to use an irrational? I am not entirely sure... Any hints/insight would be most appreciated.