Context and definitions:
Say $(X,d)$ is a compact metric space, with $f: X \rightarrow X$ continuous. For each $n \in \mathbb{N}$, the metric
$d_{n}(x,y) = \max_{0 \leq k \leq n-1}d(f^{k}(x),f^{k}(y))$
measures the maximum distance between the first $n$ iterates of $x$ and $y$. Let $\epsilon > 0$. A subset $A \subseteq X$ is $(n,\epsilon)$-$separated$ if any two distinct points in $A$ are at least $\epsilon$ apart in the $d_{n}$ metric.
My questions is about the statement: "Any $(n,\epsilon)$-separated set is finite."
My work so far:
Suppose $A \subseteq X$ is $(n,\epsilon)$-separated. Each point of $A$ is isolated in $A$, thus $A$ is closed. Since we are in a compact space, $A$ is compact. Since the space is a metric space, $A$ is bounded. My intuition is that this, along with $A$ being $(n,\epsilon)$-separated, means that $A$ is finite. This looks to be true in some simple spaces(compact subsets of $\mathbb{R}^{2}$, for example), but I'm afraid that for a space that has too many "degrees of freedom," this might not be true.. Thanks for any advice.
The easiest way to prove it, I think, is to observe that the metric $d_n$ induces the given topology on $X$.
The fact that $d_n$ is a metric is easy to verify, and obviously $d_n \geqslant d_1 = d$, hence the topology induced by $d_n$ is finer than the one induced by $d$.
But, given any $\varepsilon > 0$ and $x \in X$, the $d_n$-ball around $x$,
$$U_\varepsilon(x) = \{ y \in X : d_n(x,y) < \varepsilon\} = \bigcap_{k=0}^{n-1} \bigl(f^k\bigr)^{-1}\bigl(B_\varepsilon(f^k(x))\bigr) $$
is a neighbourhood of $x$ (in the topology induced by $d$) by the continuity of $f$. Thus the topology induced by $d$ is also finer than the one induced by $d_n$.
Thus you have a discrete closed set in the compact metric space $(X, d_n)$, which hence is finite.