Some homework questions about a Lipschitz function (cauchy sequence) Do you want to help me with my homework? The exercise is as follows:

Consider a Lipschitz function, $h:\mathbb{R}\rightarrow\mathbb{R}$, satisfying for every $x, y$:
  $$\left| h(x)-h(y) \right| \leq \alpha \left| x-y \right|$$
  with $0 \lt \alpha \lt 1$
  
  
*
  
*Show that there exists at most one $x$, with $h(x)=x$.
  
*Prove that $h$ is uniformly continuous.
  
*Take some $x_1\in\mathbb{R}$, and define inductively the sequence $(x_n)$ as $$ x_{n+1}=h(x_n), \quad  n= 1, 2, \cdots$$ Show that for every $x_1$ the sequence $(x_n)$ is a Cauchy sequence.
  
*Take some $x_1$. Define $x= \lim x_n$. Show that  $x=\lim{x_n}$ satisfies $h(x)=x$
  
*Show (using the first part), that the limits of the sequences $(x_n)$ for all choices of $x_1$ are all the same.
  

My work until now:
Part 1
Suppose $h(x_1)=x_1, h(x_2)=x_2$, and $x_1 \not = x_2$. Then, following the condition, 
$$\left| h(x_1)-h(x_2) \right|= \left| x_1-x_2 \right| \le \alpha \left|x_1-x_2\right|.$$
This means that $\frac{\left|x_1-x_2\right|}{\left|x_1-x_2\right|}=1 \le \alpha$. But $\alpha <1$ was given, so $x_1 = x_2$. There exists at most one $x$ with $h(x)=x$
Part 2
Let $\epsilon>0$ be given and choose $\delta=\frac{\epsilon}{\alpha}$. Then, for any $x, y$ with $\left|x-y\right|<\delta = \frac{\epsilon}{\alpha}$, I have
$$\left| h(x)-h(y) \right| \le \alpha \left|x-y\right| \lt \alpha\left(\frac{\epsilon}{\alpha}\right)=\epsilon,$$
which shows that $f$ is uniformly continuous.
Part 3
I don't know where to start.
Part 4

(3.) Suppose $m>n$, and consider $x_0$ instead of $x_1$.
Then $|x_n-x_m| =|f^{(\circ n)}(x_0)-f^{(\circ n)}(x_{m-n}) | < \alpha^n\cdot |x_0 - x_{m-n}|$ may help.
Note also that $|x_0-x_2|\le |x_0-x_1|+|x_1-x_2|<(1+\alpha)|x_0-x_1|$, and $|x_0-x_3| \le |x_0-x_1|+|x_1-x_3|<(1+\alpha(1+\alpha))|x_0-x_1|$, so 
$$|x_0-x_d|<(1+\alpha+\alpha^2+..\alpha^{d-1})|x_0-x_1|< \frac1{1-\alpha}|x_0-x_1|$$
(4.) Use that $h$ is continuous, hence preserves limit.
(5.) Follows from these, using (1.).
