Lemma: Let $B(a,r)$ be a ball in Banach space $X$ and $\phi$ be a contraction ($d(\phi(x),\phi(y)\leq qd(x,y),0<q<1$) from $B(a,r)\to X $. Then the function:$$ f(x)=x+\phi(x)$$ is a homeomorphism some open set $U$ from $B(a,r)$ to $B(f(a),(1-q)r)$.
Now a homeomorphism is a bijective function that is continuous and it's inverse is continous. We need to prove this. The proof starts: (Since it's a Banach space the metric is defined by the norm) $$d(f(x),f(y))=\|f(x)-f(y)\|=\|x+\phi(x)-y-\phi(y)\|\geq \|x-y\|-\|\phi(x)-\phi(y)\|\geq(1-q)\|x-y\|\implies$$"Injective". Here $\|a+b\|\geq\|a\|-\|b\|$ is used. We have to prove $y \in B(f(a),B(1-q)r)$ $f(x)=y, y=x+\phi(x),x=y-\phi(x)$ lets have a function: $g_y(t)=y-\phi(t)$ then$ \|g_y(t)-g_y(s)\|\leq q(t-s)$ we have to show: $g_y(B(a,r))\subseteq B(a,r);z\in g_y(B(a,r))\implies g_y(t)=z, t\in B(a,r). z=y-\phi(t);$ $$d(z,a)=\|y-\phi(t)-a\|=\|y-\phi(a)-a+\phi(a)-\phi(t)\|\leq \| y-f(a)\|+\|\phi(a)-\phi(t)\|<(1-q)r+qr=r $$aka $d(z,a)<r\implies z\in B(a,r)$.. I can type out the rest of the proof if need be. What is unclear to me is one of the last inqualities :$\|y-f(a)\|\leq(1-q)r ??$