If $f$ is a contraction then it is continuous. The proof might seem intuitive if just has one or more jump points which have a distance $d$ from each other.
But I am struggling, with the following problem:
If f is not continuous, such that for all intervals: $$[a,b], \text{with } a < b < c < d: f \text{ is not continuous in } [b,c]$$
where $a,d$ are constant and $b,c$ freely selectable.
Can it then be a contraction?
Edit:
$$\exists a,d: \forall b,c: (a < b < c  <d \implies f \text{ is not continuous in } [b,c])$$
I would prefer to use the definition of continuity using the epsilon-delta-criteria. However, I am fine with every (widely know) equivalent definition.
 A: Welcome to MSE!
Hint:
If $f$ is a contraction, then $d(fx,fy) < d(x,y)$ for every $x,y$.
To show $f$ is continuous, we want to show we can make $d(fx,fy) < \epsilon$ small by controlling $d(x,y)$... Of course, this is exactly the flavor of control that contractibility buys us.
Formally, say $\epsilon > 0$. We want to find a $\delta$ so that whenever $d(x,y) < \delta$, we're guaranteed $d(fx,fy) < \epsilon$...
By contractibility, we're guaranteed $d(fx,fy) < d(x,y) < \delta$.
Do you see where to go from here? What's a good choice of $\delta$?

I hope this helps ^_^
A: Let $c: (M,d) \to  (M,d)$ with $d(c(x_1),c(x_2)) \leq \gamma d(x_1,x_2)$ with $\gamma \in [0,1)$ be a contraction.
If $c$ is continuous, then
$$\forall x_0 \in M: \forall \epsilon > 0: \exists \delta > 0: c(B(x_0, \delta)) \subset B(c(x_0),\epsilon)$$
Choose $\delta = \epsilon$
Then: $c(B(x_0, \delta)) = c(B(x_0, \epsilon)) \subset  B(c(x_0), \gamma\epsilon) \subset B(c(x_0),\epsilon)$
Therefore all contractions are continous.
