Prove that mapping $f$ has fixed point

Question

Problem: Give metric $(X,d)$ is complete, $x_0 \in X$, $f:B(x_0,r) \to X$ is contraction mapping with coefficient $q<1$. Prove that if $d(f(x_0),x_0) < (1-q)r$ then $f$ have fixed point.

My attempt:

I prove this problem by using Banach's fixed point theorem. So, I need to prove 2 things below

1. $f(B(x_0,r)) \subset B(x_0,r)$.

For $x \in B(x_0,r)$, we have $$d(f(x),x_0)\le d(f(x),f(x_0))+d(f(x_0),x_0)< qd(x,x_0)+(1-q)r< qr+(1-q)r=r.$$ So, $f(x) \in B(x_0,r)$. This yields $f(B(x_0,r)) \subset B(x_0,r)$.

2. $(B(x_0,r),d)$ is complete.

This is the part makes me thing my direction is wrong because the fact is $(B(x_0,r),d)$ is not always complete.

Answer 1

Consider the closed ball of radius $r(1-q)$ around $x_0$. This is a complete metric space and it is easy to check that $f$ maps this set into itself and that it is a contraction.