If $f:(A,\rho)\to(A,\rho)$ is defined on all of $A$ and there is a constant $\alpha\in(0,1)$ such that $\rho(f(u),f(v))\leq\alpha\rho(u,v)$ for all $(u,v)\in A\times A$ so I have not understood the connection of the proof of the contraction Mapping theorem to this statement. The contraction mapping theorem states: If $f$ is a contraction of a complete metric space $(A,\rho)$, then the equation $f(u)=u$ has a unique solution. The proof makes use of the following axiom $\rho(u,v)=\rho(f(u),f(v))$, how can this be true? How can $\rho(u,v)=0$? And $u=v$? I have stuck on this for two days. Thanks for reading
The part of the proof you're asking about only shows the uniqueness of the fixed point. If there is no fixed point, the uniqueness is vacuously true and if there is only one we are done.
So, one needs to show that if $u,v\in A$ are two fixed points, then $u=v$. By definition this means $f(u)=u$ and $f(v)=v$. It follows that $\rho(f(u),f(v)) = \rho(u,v)$. The axioms of a metric space imply that if $\rho(u,v)=0$, then $u=v$. Suppose $\rho(u,v)\neq 0$. Since $\alpha<1$, this yields $$\rho(u,v)=\rho(f(u),f(v))\le \alpha \rho(u,v)<\rho(u,v),$$ which is a contradiction. Thus, $\rho(u,v)=0$ and $u=v$ as desired.