On the non-wandering set of a map Let $f: X \to X$ be a homeomorphism from a compact metric space $X$ to itself.
$x$ is said to be a non-wandering point of $f$ if for all open neighborhood of $x$ say $U$ there exists $n \in \mathbb{Z}_{\geq1}$ such that $f^n(U) \cap U \neq \emptyset$. The set of all non-wandering points of $f$ is denoted by $\Omega(f)$. I'm trying to prove the following exercise from Shub's Global Stability of Dynamical Systems (p.5,Exr.1.1):

Show that if $ x \in \Omega(f)$ and $U$ is a neighborhood of $x$, there is a sequence of integers $n_i$ tending to infinity such that the intersection $f^{n_i}(U) \cap U$ is non-empty.

There is a proposition in the book (p.2,Cor.1.4) which says if $U$ is a neighborhood of $\Omega(f)$ then for every $x \in X$ there is $ N>0$ such that for every $n \geq N $ , $f^n(x) \in U$.
May I use the proposition above to prove the main problem?
 A: The proposition you are citing says roughly that each orbit comes arbitrarily close to the nonwandering set; it does not seem to be very useful for the purposes of the exercise. Below is an argument.

First let's set up some notation and definitions. Let $X$ be compact metrizable, $f:X\to X$ be a homeomorphism. A point $x\in X$ is called $f$-nonwandering if
$$\forall U\in\operatorname{Nbhd}(x),\exists n=n(U)\in\mathbb{Z}_{\geq1}: f^n(U)\cap U\neq\emptyset.$$
(Here $\operatorname{Nbhd}(x)$ is the collection of open neighborhoods of $x$.)
Let us denote by $\operatorname{NW}(f)$ the set of all $f$-nonwandering points of $X$.
Observation: Note that $f^n(U)\cap U = f^n(U\cap f^{-n}(U))=f^n(f^{-n}(U)\cap U)$; thus
$$f^n(U)\cap U\neq\emptyset \iff f^{-n}(U)\cap U\neq\emptyset.$$
Thus the sign of the time $n$ when the orbit of $U$ hits $U$ again is not important, that is,
$$x\in \operatorname{NW}(f) \iff \forall U\in\operatorname{Nbhd}(x),\exists n=n(U)\in\mathbb{Z}\setminus0: f^n(U)\cap U\neq\emptyset.$$
(It seems in your definition there is a typo; $n=0$ mustn't be allowed.)

Claim: We claim the following:
$$x\in\operatorname{NW}(f) \iff \forall U\in\operatorname{Nbhd}(x),\forall N\in\mathbb{Z}_{\geq1},\exists n=n(U,N)\in\mathbb{Z}_{> N}: f^n(U)\cap U\neq\emptyset.$$
Proof: ($\Leftarrow$) is clear. For $(\Rightarrow)$, let $x\in\operatorname{NW}(f)$ and fix $U\in\operatorname{Nbhd}(x)$ and $N\in\mathbb{Z}_{\geq 1}$. If $x$ is an $f$-periodic point, say with $f^p(x)=x$ with $p\in\mathbb{Z}_{\geq1}$, then $f^{pN}(U)\cap U\ni x$. So we may assume that $x$ is not an $f$-periodic point. Then the orbit of $x$ is infinite and in particular
$$x,f(x),f^2(x),...,f^{N}(x)$$
are all distinct. Thus for any $i\in\{1,2,...,N\}$ there are open $A_i\ni x$ and $B_i\ni f^i(x)$ such that $A_i\cap B_i=\emptyset$ (because $X$ is Hausdorff). Then for each $i$,  $C_N=\bigcap_{i=1}^N (A_i\cap f^{-i}(B_i))\ni x$ is an open neighborhood of $x$ and $f^i(C_N)\cap C_N\subseteq B_i\cap A_i=\emptyset$. Put $U_N=U\cap C_N\ni x$. Then we have that $U_N$ is an open neighborhood of $x$ and the open subsets
$$U_N, f(U_N), f^2(U_N),..., f^N(U_N)$$
are all pairwise disjoint. Since $x$ is an $f$-nonwandering point there is a number $n\in\mathbb{Z}_{\geq1}$ such that $\emptyset\neq f^n(U_N)\cap U_N\subseteq f^n(U)\cap U$. By construction $n>N$.

As a final note, the above observation also gives the following:
$$x\in\operatorname{NW}(f) \iff \forall U\in\operatorname{Nbhd}(x),\forall N\in\mathbb{Z}_{\geq1},\exists n=n(U,N)\in\mathbb{Z}: |n|>N \text{ and } f^n(U)\cap U\neq\emptyset.$$
