Contraction Principle: Existence of a fixed point proof. Definition of contraction: 
A function $f : M \mapsto M$ is a contraction $\iff$ $\exists r \in (0, 1)$ such that $d\left( f(x), f(y) \right) \leq r d(x, y)$ $\forall$ $x, y \in M$.
Statement of the theorem:
Let $M$ be a complete metric space and $f : M \mapsto M$ be a contraction. Then, there exists a unique fixed point of $f$ (i.e. $\exists$ $x_0 \in M$ such that $f(x_0) = x_0$). Moreover, $x_0 = \lim \limits_{m \to \infty} f^m(x)$ $\forall x \in M$.

Notation: $f^m = \underbrace{f \circ f \circ \cdots \circ f}_{\text{m times }}$
I just need help with the existence part. I already showed for any $x \in M$, $\left( f^m(x) \in M \right)_{m \in \mathbb{N}}$ is a Cauchy sequence and hence convergent by completeness of $M$.
From this, how do I get to $f(x_0) = x_0$? 
 A: Since $f$ is a contraction, $f$ is uniformly continuous, for if $\varepsilon>0$ is arbitrary, and if $x$ and $y$ are two points of $M$ such that $d(x,y) < \delta := \varepsilon/r$ then
$$d(f(x),f(y)) \leq r \cdot d(x,y) < \varepsilon.$$
With this, follows the desired result:
$$\begin{align}
f(x_0) = f\Big( \lim_{m\to\infty} f^m(x) \Big) 
&\stackrel{(1)}{=} \lim_{m\to\infty} f(f^m(x)) \\
&= \lim_{m\to\infty} f^{m+1}(x) = x_0
\end{align}$$
where, in $(1)$, we use the fact that $f$ is continuous.
A: To prove the existence, do similar work as you did.


*

*Take any $x\in M$, and $x_1 = x$.

*Inductively, $x_{n+1} = f(x_n)$ for $n\geq 1$.


Then by the contraction property, the sequence $(x_n)$ is Cauchy.
By the completeness of $M$, there is $x_0 = \lim x_n$.
To prove the uniqueness, assume there are two distinct fixed points $x_0$ and $y_0$.
Then $d(f(x_0),f(y_0)) = d(x_0,y_0)$, which contradicts to the contraction property.
A: It's easy to see that $f$ is continuos funtion
Consider $\{x_n\}_{n\ge 0}$: $x_0=x$ and $x_{n+1}=f(x_n)$
We found that $d(x_m,x_n)\le rd(x_{m-1},x_{n-1})$ (1)
So with $\epsilon >0$, we choose $n>log_r(\frac{(1-r)\epsilon}{d(x_1,x_0})$ then $\forall p\in\mathbb{N}$: 
$d(x_{n+p},x_n)$
$\le d(x_{n+p},x_{n+p-1})+d(x_{n+p-1},x_{n+p-2})+...+d(x_{n+1},x_n)$ 
$\le (r^{n+p-1}+r^{n+p-2}+...+r^n)d(x_1,x_0)$
=$r^n\frac{1-r^p}{1-r}d(x_1,x_0)$
$\le r^n\frac{d(x_1,x_0)}{1-r}$
$\le \epsilon$
So $\{x_n\}_{n\ge 0}$ is a Cauchy sequence, in complete metric space $M$ it converges at $N$
Then $N=lim_{n\rightarrow \inf}x_n=\lim_{n\rightarrow \inf}x_{n+1}=\lim _{n\rightarrow}f(x_n)=f(N)$, by the continuty of $f$
