Question: Let $(X,d)$ be a metric space so that for every $x \in X$ and every $r>0$, the closed ball $\overline{N(x,r)} = \left \{ y\in X: d(y,x) \leq r \right \}$ is compact. Let $f:X \to X$ be a homeomorphism. For every $n \in \mathbb{N}$ define $f^n = f \circ ... \circ f$, $n$-times. Similarly, define $f^{-n} = f^{-1} \circ ... \circ f^{-1}$, $n$-times. Define $f^0 = id$ ($f^0 (x) = x$ for any $x \in X$). Let $x \in X$ and suppose $F_x = \left \{ f^n (x) : n \in \mathbb{Z} \right \}$ is a closed subset of $X$. Prove that $F_x$ is a discrete subset of $X$.
Attempt: Suppose for contradiction that $F$ is not discrete. Then there exist $n \in \mathbb{Z}$ so that $f^n (x)$ is not an isolated point. So, $f^n (x)$ is a limit point of $F_x$. It follows that there exist $n_1,n_2,...$ such that $f^{n_1} (x), f^{n_2} (x), ...$ converges to $f^n (x)$ and $f^{n_m} (x) \neq f^n (x)$. So, $f^n (x) = \lim_{m\to \infty} f^{n_m} (x)$. Then $f^k \left( f^n (x) \right) = \lim_{m\to \infty} f^k \left( f^{n_m} (x) \right) \Leftrightarrow f^{k+n} (x) = \lim_{m\to \infty} f^{k+ n_m} (x)$. It follows that every point of $F_x$ is a limit point. So, $F_x$ is a perfect set.
This is where I am stuck. Is my current approach correct, or is there a more elegant (perhaps easier) way of doing this? I know that perfect sets in $\mathbb{R}^n$ are uncountable, and that the set $F_x$ is countable, but I am not sure on how to proceed. (I think that the compact closed ball condition should be used soon to obtain the contradiction that I am not sure on how to achieve.) Also, for clarification, I am trying to stay with using Analysis (I am using baby Rudin) and will clarify some definitions:
Definition: $f$ is a homeomorphism if $f$ is a bijection, is continuous, and has a continuous inverse.
Definition: A set $E$ is perfect if $E$ is closed and if every point of $E$ is a limit point of $E$.
Definition: A subset $Y \in X$ is discrete if for every $y \in Y$ there exists some $r_y >0$ so that $N(y,r_y) \cap Y = \left \{ y \right \}$.