I tried to prove something but I could not, I don't know if it's true or not, but I did not found a counterexample.
Let $(a_n)$ be a sequence in a general metric space such that for any fixed $k \in \mathbb N$, we have $|a_{kn} - a_n| \to 0$. Is $a_n$ necessarily a Cauchy sequence?
It need not be a Cauchy sequence; here’s a counterexample in $\mathbb{R}$.
Let $a_n = \ln\ln n$. Then
$$\begin{align*} |a_{kn} - a_n| &= \ln\ln kn - \ln\ln n \\ &= \ln\frac{\ln kn}{\ln n} \\ &= \ln\frac{\ln k + \ln n}{\ln n} \\ &= \ln\left(1 + \frac{\ln k}{\ln n}\right) \\ &\to 0 \end{align*}$$
as $n \to \infty$, but the sequence $\langle a_n \rangle_n$ is unbounded and hence not Cauchy.
The additional question about adding the condition that the sequence is bounded caught my attention. (It is in Daniel's comment under Brian's answer.)
First I was able to get an example only for $k=2$:
$a_{2^x+y}=\frac y{2^x}$ for $0\le y<2^x$.
When trying to find an example working for all $k$'s I got stuck for some time, so I tried to modify Brian's example:
$a_n=\sin\left(\frac\pi2 \lg\lg n\right)$.
To show that it fulfills the requirements, we can use $|\sin x-\sin y|\le |x-y|$ and the same reasoning as Brian did. (In detail: $|a_{kn}-a_n| = |\sin(\frac\pi2\lg\lg kn)-\sin(\frac\pi2\lg\lg n)| \le \frac\pi2(\lg\lg kn - \lg\lg n)$, an the RHS converges to 0, as shown in Brian's answer. Thus $|a_{kn}-a_n|\to 0$.)
But this sequence has a subsequence convergent to 0 (for $n=2^{2^{2k}}$) and a subsequence convergent to 1 (for $n=2^{2^{4k+1}}$). Hence it is not convergent and, consequently, not Cauchy.
I hope I did not miss some mistake there.
