Prove that $f(x) = \cos(x)$ has a unique fixed point. The following question comes from Chapter 6.4, Exercise 4 on page 156 in the set of notes Topology Without Tears.

Using Exercise 2 and 3, show that while $f: \mathbb R \to \mathbb R$ given by $f(x)=\cos(x)$ does not satisfy the condition of the Contraction Mapping Theorem, it nevertheless has a unique fixed point.

The "Exercise 2 and 3" mentioned in the question (which I've already done) were


  
*Extend the Contraction Mapping Theorem by showing that if $f$ is a mapping of a complete metric space $(X,d)$ into itself and $f^N$ is a contraction mapping for some positive integer $N$, then $f$ has precisely one fixed point.
  
*Use the Mean Value Theorem to prove the following:
Let $f: [a,b] \to [a,b]$ be differentiable. Then $f$ is a contraction if and only of there exists an $r \in (0,1)$ such that $|f'(x)|\le r$, for all $x \in [a,b]$.

Can anybody please show me how to use those to show that $f(x)=\cos(x)$ has a unique fixed point?
I know, from a Numerical Analysis course, that this fixed point can be found using a method such as Newton's Method, but I am not sure how to prove that it actually exists (and is unique).

This isn't the suggested proof at all, but it's a lot simpler, and I hope it's of some value. @Jonas Meyer's comment seems to give a good route to solving the problem as suggested, so there's no reason to duplicate. 
A fixed point of $x \mapsto \cos x$ must be between $-1$ and $1$, because at a fixed point $\cos x = x$, and the codomain of $\cos$ is $[-1, 1]$. 
For $-1 < x < 0$, cosine is positive but $x$ is negative, so we need only consider $0 \le x \le 1$. 
At $0$, $g(x) = \cos x - x$ is positive. At $1$, $g(1)$ is negative (by direct evaluation). Throughout the interval, $g'(x) < 0$, so $g$ has at most one zero, say $b$,  in the interval, and at that point $\cos b - b = 0$, so $\cos b = b$. 
