Q:

If $(E,d)$ is a compact metric space, $x_1 = f(x_0)$ and $x_{n+1}=f(x_n)$, then $x_0$ is an accumulation point.

Let $(E,d)$ be a compact metric space. And let $f: E \rightarrow E$ such that $$\forall x,x' \in E, d(f(x),f(x')) = d(x,x') $$
  and  $$x_1 = f(x_0), \forall n \in \mathbb{N}, x_{n+1}=f(x_n)=f^{n+1}(x_0) $$
  Show that $x_0$ is an accumulation point of $(x_n)_{n\in \mathbb{N}}$.

Now I see two ways to do it: either I show that there exists $(x_{\psi(n)})_{n \in \mathbb{N}}$ that converges to $x_0$, or I show that for any $\epsilon >0 $ and for any $N \in \mathbb{N}$, there exists an $n >n$, such that $d(x_n,x_0)<\epsilon$. As the space is compact, I think I should approach the first method. 
I kind of see that for any $n\in \mathbb{N}$, we have $d(x_n,x_o)\leq d(f(x_{n-1}),f(x_{n-2}) )+ d(f(x_{n-2}), x_0) = d(x_{n-1},x_{n-2}) + d(f(x_{n-2}), x_0)$ and etc, and I could work this out until coming to an inequality that would have on its right hand something having $d(x_1, x_0)$, but I am rather unsure how to work when $epsilon$ is very small. I am actually unsure if this helps.

A:

Since $E$ is compact, you can extract a Cauchy sequence $(x_{u(n)})$ for every $c>0$, there exists $N_c$ such $n>m>N_c$ implies that $d(x_{u(n)},x_{u(m)})<c$ this implies that $d(f^{u(n)}(x_0),f^{u(m)}(x_0))=d(f^{u(n)-u(m)}(x_0),x_0)=d(x_{u(n)-u(m)},x_0)<c$. We deduce that $x_0$ is an accumulation point.

