Contraction maps should be considered together with Banach fixed-point theorem. So ...
What does it mean to be a contraction mapping in the context of the
sequence of real numbers given by $f(x_n)=x_{n+1}$?
As others stated $d(x,y)= |x-y|$ is a metric.
And what does it tell us about such sequence?
Well, it tells that the sequence is Cauchy. I.e.
$$|x_{n+1}-x_{n}|=|f(x_{n})-f(x_{n-1})|\leq k |x_{n}-x_{n-1}|\leq ... \leq k^{n}|x_1-x_0|$$
and with this in mind:
$$|x_{n+p}-x_n|=|x_{n+p}-x_{n+p-1}+x_{n+p-1}-x_n|=|x_{n+p}-x_{n+p-1}+...+x_{n+1}-x_n|\leq \\
|x_{n+p}-x_{n+p-1}|+...+|x_{n+1}-x_n|
\leq k^{n+p-1}|x_{1}-x_{0}|+..+k^{n}|x_{1}-x_{0}|=\\
|x_{1}-x_{0}|\left(k^{n+p-1} + ... +k^{n}\right)=k^{n}|x_{1}-x_{0}|\left(k^{p-1} + ... +1\right)<...$$
with the last term, we can add all the terms of the geometric progression because $0\leq k < 1$ and
$$... < k^{n}|x_{1}-x_{0}|\sum_{t=0}^{\infty}k^t=\frac{k^n}{1-k}|x_{1}-x_{0}| \rightarrow 0, n\rightarrow \infty$$
But, we know that every Cauchy sequence in a complete metric space has a limit and every compact is also complete. Also, every segment of the form $[\alpha, \beta] \subset \mathbb{R}$ is compact.
Here comes Banach fixed-point theorem saying that if $f:[\alpha, \beta] \rightarrow [\alpha, \beta]$ is a contraction mapping then it admits a unique fixed point $f(x^*)=x^*$ and for $\forall x_0 \in [\alpha, \beta]$ the sequence constructed like $x_{n+1}=f(x_n)$ has $x^*$ as its limit. From this perspective
Contraction mapping in the context of $f(x_n)=x_{n+1}$.
is a way to check if the sequence converges.
