A homeomorphism $f:X\rightarrow X$ is called expansive if there exists a constant $\delta>0$ such that if $d(f^n(x),f^n(y))<\delta$ for all $n\in\mathbb{Z}$ then $x=y$.
How to prove that if $X$ is compact, this property is topological, i.e. depends not on the metric $d$, but on the topology?
Suppose $d$ and $d^\prime$ are metrics that both induce the topology $\tau$ on the compact space $(X,\tau)$. Suppose further that $f : X \to X$ is expansive with regard to $d$ with expansivity constant $\delta > 0$. $X$ is compact, so we can cover it with a finite amount of open balls of radius $\delta / 2$ in the metric $d$. We will denote this open cover of X by $$\mathcal{U} = \lbrace B_d^{i}(x_i, \delta /2) \mid 1 \leq i \leq n \rbrace. $$ $X$ is compact and $\mathcal{U}$ is an open cover of $X$, so by the Lebesgue number lemma there is a $\delta^{\prime} > 0$ such that every subset of $X$ having diameter less than $\delta^{\prime}$ in the metric $d^{\prime}$ is contained in one of these balls.
Thus if $d^{\prime}(x,y) < \delta^{\prime}$, we have that $x,y \in B_d^{i}(x_i, \delta /2) \textrm{ for some } 1 \leq i \leq n $, from which it follows by the triangle inequality that $$d(x,y) \leq d(x, x_i) + d(x_i, y) < \delta / 2 + \delta / 2 = \delta $$
Hence if we have $d^{\prime}(f^{n}(x), f^{n}(y)) < \delta^{\prime}$ for every $n \in \mathbb{Z}$, we also have $d(f^{n}(x), f^{n}(y)) < \delta$ for every $n \in \mathbb{Z}$, so $x = y$ by the assumption. Therefore $f$ is expansive also with regard to $d^{\prime}$, with expansivity constant $\delta^{\prime}$.
This shows that the expansivity of $f$ does not depend on the particular choice of metric.