Show that $f$ is a strong contraction when $f$ is continuously differentiable. 
Let $f: [a,b] \to R$ be a differentiable function of one variable such that $|f'(x)| \le 1$ for all $x\in [a,b]$. Prove that $f$ is a contraction. (Hint: use MVT.) If in addition $|f'(x)| < 1$ for all $x \in [a,b]$ and $f'$ is continuous, show that $f$ is a strict contraction.

Using MVT, $|f(x) - f(y)| = |f'(c)(x-y)| \le |x-y|$ for $c$ between $x$ and $y$.
I don't know the proof for a strict contraction. I guess that I need to use the continuity of $f'$, but I am not sure how to use it. Any help would be appreciated.
 A: You're almost there! A (weak) contraction is defined as a function $f: A \to \mathbb{R}$ such that $$|f(x) - f(y)| \leq k  | x - y| ~\forall x,y \in A,\:\:\: 0 \leq k \leq 1$$ (in your case $A = [a,b]$.)
This might look a little familiar, as you have already concluded that $|f(x) - f(y)| < |x - y|$ (by removing the middle inequality). We just need to make sure that $k = |f(x)|$ meets the required bounds. Combine $0 \leq |f(x)|$ (a property of absolute value) with the given $|f(x)| \leq 1$ for all $x \in [a,b]$, you can conclude $0 \leq |f(x)| \leq 1$, so $1\geq k \geq |f(c)| \geq 0$ (by continuity and $c \in[a,b]$).
So you have effectively shown that $f$ is a weak contraction.
Showing that if $|f(x)| < 1$, then $f$ is a strict contraction follows a similar path. @bitesizebo is correct in noting that $f'$ continuous is essential here. A strict contraction replaces the weak inequality with a strong inequality. A function is a strict contradiction if
$$|f(x) - f(y)| < k  | x - y| ~\forall x,y \in A,\:\:\: 0 \leq k \leq 1$$ (in your case $A = [a,b]$.)
You know that $|f(x) - f(y) \leq |f(c)(x-y)|$ by MVT. You can separate the RHS of the equation and get $|f(c)(x-y)| = |f(c)||x-y|$. Using the fact that $f'$ is continuous and $|f'(x)| < 1$, you need to prove that $|f'(c)| < 1$ (level of detail here depends on your class), as @bitesizebo says.
There are a few ways to show that $f$ satisfies $\sup_{[a,b]} |f'(c)| < 1$. My favorite method is a bit more broad than @Michael Hardy's approach and proves the Extreme Value Theorem along the way. I use prefer the 'fact' that the image of a compact set under a continuous function is also compact. This works in any topological space, not just $\mathbb{R}$. So $f'([a,b])$ is a closed set, so it must attain all of it's limit points (the limit of any sequence of numbers in $f'([a,b])$ must also be a point in $f'([a,b])$). So the infimum and supremum of $f'([a,b])$ must be obtained and be in the set. So as $|f'(c)| < 1$ for all $c \in[a,b]$, you must have also hit the supremum and infimum of $f'([a,b])$. So $\sup_{[a,b]}|f'(c)| < 1$. You can fill in the algebraic details and inequalities.
Once you have $\sup_{[a,b]}|f'(c)| < 1$, you can plug this in to the statement we got from the MVT, and conclude $|f(x) - f(y) < k |y-x|$ where $|f'(c)| < \sup_{[a,b]}|f'(c)| \leq k < 1$ (I've never seen the assumption $k>0$ used to define a strict contraction since it could give a 'strongest' possible contraction step, although I don't have my copy of Rudin handy), and conclude that $f$ is a contraction map.
A: Besides the mean-value theorem, another "MVT" is the "maximum-value theorem": A continuous function on a closed bounded interval has an absolute maximum. If $|f'|$ is continuous, then there is some point $c\in[a,b]$ for which $|f'|$ is at least as big as it is at any point in that interval. And that value must be less than $1,$ by hypothesis. Then apply the mean value theorem again.
(You probably won't see that other theoem called the "MVT", for "maximum-value theorem", since it is instead called the extreme-value theorem, applying to both maxima and minima.)
