Prove the Contraction Mapping Theorem. Prove the Contraction Mapping Theorem.

Let $(X,d)$ be a complete metric space and $g : X \rightarrow X$ be a map such that $\forall x,y \in X, d(g(x), g(y)) \le \lambda d(x,y)$ for some $0<\lambda  < 1$. Then $g$ has a unique fixed point $x^* \in X $, and it attracts everything, i.e. for any $x_0 \in X$ , the sequence of iterates $x_0, g(x_0), g(g(x_0))$, ... converges to the fixed point $x^* \in X$.

The hint I am given are for existence and convergence - prove that the sequence is Cauchy. For uniqueness, choose two fixed points of $g$ and apply the map to both.
Still a bit do not know how to proceed after looking at the hint. Could anyone help me based on those hints?
 A: From 
$$d(x_{n+1},x_n)=d(g(g(x_{n-1})),g(x_{n-1}))\le \lambda d(g(x_{n-1}), x_{n-1})=d(x_n,x_{n-1})$$
we get after $n$ applications of that inequality
$$d(x_{n+1},x_n)\le \lambda d(x_n,x_{n-1}) \le \lambda^2 d(x_{n-1},x_{n-2}) \le \cdots\le \lambda^n d(x_1,x_0)\tag{1}$$
Now we want to show that $(x_n)_n$ is a Cauchy sequence. So let $\epsilon>0$.
We assume $x_1\not=x_0$ (otherwise $x_0$ is already a fixed point). Set $c=d(x_1,x_0)>0$. 
Since $0<\lambda<1$, the sum $\sum_{n=0}^\infty \lambda^n$ converges (to $1/(1-\lambda)$). Therefore we can pick $N$ large enough such that
$$\sum_{k=n}^\infty \lambda^k<\frac{\epsilon}{c}$$
for all $n\ge N$.
Then for $m>n\ge N$ we have by the triangle inequality
$$d(x_m,x_n)\le \sum_{k=n}^{m-1} d(x_{k},x_{k+1})$$
Applying $(1)$ we obtain
$$d(x_m,x_n)\le c\sum_{k=n}^{m-1} \lambda^k\le c\sum_{k=n}^\infty \lambda^k<c\cdot\frac{\epsilon}{c}=\epsilon$$
So $(x_n)_n$ really is a Cauchy sequence. Since $(X,d)$ is complete, it converges to a limit $x\in X$.
By the equation $x_{n+1}=g(x_n)$, the limit satisfies $x=g(x)$, so it is a fixed point.
Uniqueness is trivial, let $y$ be another fixed point of $g$. Then
$$d(x,y)=d(g(x),g(y))\le \lambda d(x,y)$$
If now $x\not=y$, then $d(x,y)>0$, so we can divide by $d(x,y)$ to obtain $\lambda\ge 1$, a contradiction. Therefore, $x=y$.
A: One proviso: You do need that $X$ is non-empty in the statement. 
To elaborate on the hints you have been given;
To prove existance, pick $x_0 \in X$, and call (for convenience of writing) $g(x_0) = x_1, g(x_1) = x_2$ etcetera. Let $d(x_0,x_1) = d$. Then $d(x_1,x_2) \leq \lambda d$, and $d(x_2,x_3) \leq \lambda^2 d$...can you see how to extend this to show that the sequence is Cauchy?
For uniqueness, suppose $x$ and $y$ are both fixed points. What is $d(g(x),g(y))$?.
