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.
A: Recall that every sequence in a compact space has a convergent subsequence. Then we have that $\{x_n\}$ has a convergent subsequence $\{x_{k_n}\}$, which converges to $x$. Note that in particular this sequence is a Cauchy Sequence, which will help us construct a sequence converging $b_n$ to $x_0$.
For $\epsilon = 1$ $\exists n_1 \in \mathbb{N}$ s.t. $n,m \ge n_1 \implies d(x_{k_n},x_{k_m}) < 1$. In particular we have that: $d(x_{k_{n_1}},x_{k_{n_1 + 1}}) < 1$ But using the property of $f$ we have:
$$1 > d(x_{k_{n_1}},x_{k_{n_1 + 1}}) = d(f^{k_{n_1}}(x_0),f^{k_{n_1+1}}(x_0)) = d(x_0,f^{k_{n_1+1}-k_{n_1}}(x_0))$$
So now choose $b_1 = f^{k_{n_1+1}-k_{n_1}}(x_0)$
Similarly for $\epsilon = \frac 12$ $\exists n_2 \in \mathbb{N}$ s.t. $n,m \ge n_2 \implies d(x_{k_n},x_{k_m}) < 1$. In particular we have that: $d(x_{k_{n_2}},x_{k_{n_2 + 1}}) < 1$. So similarly as above we have:
$$\frac 12 > d(x_{k_{n_2}},x_{k_{n_2 + 1}}) = d(f^{k_{n_2}}(x_0),f^{k_{n_2+1}}(x_0)) = d(x_0,f^{k_{n_2+1}-k_{n_2}}(x_0))$$
Now choose $b_2 = f^{k_{n_2+1}-k_{n_2}}(x_0)$
Continue doing this for $\epsilon = \frac 1n$, where $n \in \mathbb{N}$ to get a convergant sequence s.t. $b_n \to x_0$. You can be more rigorous and address the fact that the elements in $\{b_n\}$ might not appear in the same order as in $\{x_n\}$, which shouldn't be big of a problem, but would require some more tedious wording.
Note that in this manner you can prove an even more stronger result. It's enough for the metric space to be precompact, as all we need is a Cauchy subsequence of $\{x_n\}$ and a precompact space guarantees it.
