Showing the map $T: D\to D$ is a contraction I need to prove that a map $T:D \to D$ is a contraction. Most of the work has been done already, I am just stuck at trying to break apart some inequalities (see below). Here's the original problem:

Let $\alpha,\beta\in\mathbb{R}^n$, $a\in \mathbb{R}$, and $A$ be an $n\times n$ nonsingular matrix. Use the Banach Fixed Point Theorem to prove that for a sufficiently small $\epsilon > 0$, $$\alpha - Ax + \epsilon(a - \beta^tx)x = 0_v \tag{*}$$ has a unique solution in $$D = \{x \in \mathbb{R}^n \,|\, \|x - A^{-1}\alpha\| \le 1\}.$$

Here's what I've done:
We know that $D$ is a closed subset of the Banach space $\mathbb{R}^n$ because it is a closed ball centered at $A^{-1}\alpha$ with radius $1$.
We can rearrange $(*)$ to get the fixed point $$T(x) = A^{-1}[\alpha + \epsilon(a - \beta^Tx)x].$$ Also, observe that for any $x \in D$, $$\|x\| = \|x - A^{-1}\alpha + A^{-1}\alpha\| \le \|x - A^{-1}\alpha\| + \|A^{-1}\alpha\| \le 1 + \|A^{-1}\alpha\|$$
So clearly $\|x\| \le 1 + \|A^{-1}\alpha\|$. Choose $$\epsilon \le {1 \over (|a| + \|\beta\|(1 + \|A^{-1}\alpha\|))\|A^{-1}\|(1 + \|A^{-1}\alpha\|)}.$$ Then it is easy to verify that $T(x) \in D$, showing that for any $x \in D$, $T: D \to D$. 
Here's where I'm stuck.
I need to show that $T$ is a contraction. Here's what I've got:
$$\begin{align}\|T(x) - T(y)\| & = \|A^{-1}[\alpha + \epsilon(a - B^Tx)x] - A^{-1}[\alpha + \epsilon(a - B^Ty)y]\| \\ & \le \epsilon \|A^{-1}\|\|a(x - y) - [(\beta^Tx)x - (\beta^Ty)y]\|\end{align}$$
My professor gave me the hint that I need to break up the inequality so that I can get $\|x - y\|$ and anywhere that $\|x\|$ occurs to estimate it with the previously derived result $\|x\| \le 1 + \|A^{-1}\alpha\|$.
Can anyone provide a hint on how to break up this inequality to do that? Thanks.

Hint: Write $$(\beta^T x)x-(\beta^Ty)y = (\beta^T x)(x-y) + (\beta^T (x-y))y$$
