We say $(X,f)$ is expansive if there is $c(f)>0$ such that if $d(f^{n}(x), f^{n}(y))< c(f)$ for every $n\in Z$ then $y=x$. Let $(X,f)$ is expansive with constant $c(f)$ and for infinite set $\{n_k \}$, $d(f^{n_k}(x), f^{n_k}(y))< c(f).$
My question is if $x=y$? ($x,y$ are not periodic points)
Not necessarily. Let $X=\{0,1\}^\mathbb{Z}$ be the space of all bi-infinite sequences of symbols $0$ and $1$ with the metric $$ d(x,y) := 2^{-\inf\{|i|: x_i\neq y_i\}} $$ (with $\inf\varnothing:=\infty$), which induces the product topology. Let $f$ be the shift map defined by $(f(x))_i:=x_{i+1}$. Note that $d(f^n(x),f^n(y))<1$ if and only if $x_n=y_n$. So, $(X,f)$ is expansive with constant $1$. Now, choose two sequences $x$ and $y$ that are distinct but agree on infinitely many positions $\{n_k\}$.