Let $(X,d)$ be a compact metric space and let $f:X\to X$ be an isometry. We need to show that $f$ is not topologically mixing. Recall that if $Y$ is a topological space and a map $g:Y \to Y$ is continuous then $g$ is topologically mixing if for every pair of non-empty open sets $U, V \subseteq X$ there exists $n_0 \in \mathbb{N}$ such that $f^{-n}(U) \cap V \neq \emptyset \, \forall n \geq n_0.$
My attempt: Fix an arbitrary $x \in X.$ Suppose $f$ was topologically mixing and applying the mixing condition to $U= V = B(x, \frac{1}{k})$ we can find a strictly increasing sequence $(n_{k}) \in \mathbb{N}$ such that $f^{-n}(B(x, \frac{1}{k}))\cap B(x, \frac{1}{k}) \neq \emptyset \, \forall n \geq n_k$
$\Rightarrow \, \, \exists \, (x_k) \in B(x, \frac{1}{k})$ such that $d(f^{n}(x_{k}), x) < \dfrac{1}{k}$ for $n\geq n_{k}$
By the fact that an isometry of a compact metric space must be a homeomorphism and that $f$ is an isometry we can say:
$\Rightarrow d(x_k, x)< \dfrac{1}{k}$ and $d(x_k, f^{-n}(x))< \dfrac{1}{k}$ for $n \geq n_{k}$
$\Rightarrow d(x, f^{-n}(x))< \dfrac{2}{k}$ for $n \geq n_{k}$ (by the triangle inequality)
$\Rightarrow d(x, f^{n}(x))< \dfrac{2}{k}$ for $n \geq n_{k}$
Then taking limits as $k \to \infty$, we note that $n_{k} \to \infty$ and hence $n \to \infty.$
$\Rightarrow f^{n}(x) \to x$ so by continuity, $f^{n+1}(x)\to f(x)$ and we must therefore have $f(x)=x.$
But $x \in X$ was arbitrary. So $f$ must be the identity, but it's immediately clear that the identity cannot be mixing (take two disjoint non-empty open subsets of $X.$) and we get a contradiction.
Therefore no such $f$ can be mixing.
My question: Is this proof correct? If so, what might be a simpler proof perhaps avoiding use of the fact that such an $f$ is in fact invertible?
