If a self map $f$ of a metric space $X$ satisfies $d(fx,fy) \le K d(x,y)$ for $K < 1$, then by the triangle inequality $d(x,y) \le d(x,fx) + d(fx,fy) + d(fy,y)$ which gives $d(x,y) < {1 \over 1-K} (d(x,fx) + d(y,fy))$. Then if $f^n$ denotes $f$ composed with itself $n$ times, substituting $f^n(x)$ for $x$ and $f^m(y)$ for $y$ in the above inequality gives $d(f^n(x),f^m(y) < {K^n + K^m\over 1-K} d(x,fx)$, so $f^n(x)$ is Cauchy, hence if $X$ is complete it converges to a limit $x_0$ which is clearly a fixed point of $f$. This is the Banach Contraction Theorem.
(Note: we have used the obvious fact that $d(f^n(x),f^n(y)) \le K^n d(x,y)$)
