Let $X$ be a compact Hausdorff topological space without isolated point . Also, there exist open cover $\mathcal{V}=\{V_i\}_{I=1}^m$ and a homeomorphism $f:X\to X$ such that $card(\bigcap _{j\in\mathbb{Z} }(f^j(V_{k_j}))<\infty$ for all $\{k_j\}_{j\in\mathbb{Z}}\in \{1,2, \ldots, m\}^{\mathbb{Z}}$. We try to show that $X$ is metrizable, but I don't know this proof is true?
Given a point $x$ and a neighborhood $W$ of $x$, we can find $N\in\mathbb{N}$ such that $x\in\bigcap_{|j|<N}f^j(V_{k_j})\subseteq W$. This implies that the finite intersections of the elements of $\mathcal{V}$ and its iterates form a countable basis for the topology of $X$. Hence Since $X$ is compact and Hausdorff space, we can apply Urysohn’s Theorem to conclude that $X$ must be metrizable.
Please help me to know this proof is true?
The first sentence of your proof is incomplete: you do not justify the claim that the intersection is contained in $W$. The double-arrow space $A$ is a counterexample. Let $V_1$ the clopen interval $[\langle 0,1\rangle,\langle\frac12,0\rangle]$ and $V_2$ the complementary interval. Let $f:A\to A$ be defined by $f(x,i)=\langle x+\pi,i\rangle$, where addition is modulo $1$. This satisfies the condition above: All intersections have at most two points, they can be of the form $\{\langle x,0\rangle\}$, $\{\langle x,1\rangle\}$, or $\{\langle x,0\rangle,\langle x,1\rangle\}$. Yet the space is not metrizable.