Completeness respect two metric.

Question

Let be $C\subset X$ a subset of a real Banach space $(X,\|\cdot\|)$ and let be $\Sigma=\{x\in C:\|x\|=1\}$. Suppose that

1. $(C,d)$ is a complete metric space
2. $\|x-y\|\le A d(x,y) \ \ \forall x,y\in C$ and for some $A>0$
3. $d'$ is a metric on $\Sigma$
4. $\frac12d'(x,y)\le d(x,y)\le d'(x,y) \ \ \forall x,y\in \Sigma$

I want to show thad $(\Sigma,d')$ is a complete metric space.

Let $\{x_n\}$ a Cauchy sequence respect to $d'$. By (4) we have that $\{x_n\}$ is a Cauchy sequence respect to $d$ and so it exists $y\in C$ such that $d(x_n,y)\to0$.

Also by (4) we have that $d'(x_n,y)\to 0$.

We need to show that $\|y\|=1$, but from (2) we have that $\|x_n-y\|\to 0$ and so $$\|y\|=\|y-x_n+x_n\|\le\|y-x_n\|+\|x_n\|=\|y-x_n\|+1\to 1.$$

Do I miss something? Is it correct my proof?