$\textbf{The Problem:}$ Fix $0<\alpha<1$ and define $T:\mathbb R^2\to\mathbb R^2$ by $T(x,y)=(x,\alpha y)$. Prove that $T$ is not a contraction on $\mathbb R^2$ with the euclidean norm, but if we fix $x$ and let $L_x=\{(x,y):y\in\mathbb R\}$, then $T$ is a contraction on the line $L$. Identify the fixed point of $T$ on $L_x$. Also, we say that $f:\Omega\to \Omega$, where $(\Omega,d)$ is a metric space, is a contraction if $f$ is Lipschitz continuous with Lipschitz constant $\beta$ such that $0\leq\beta<1$.
$\textbf{My Thoughts:}$ Suppose for a contradiction that $T$ is a contraction on $\mathbb R^2$, and let $\beta$ be the contraction constant. If $\beta<\alpha$, pick $(0,0)$ and $(0,\beta)$, and observe that since $T$ is a contraction we have $$\sqrt{\alpha^2\beta^2}\leq\beta\sqrt{\beta^2},$$ and hence, $\alpha\leq\beta$, which is a contradiction. Next, if $\alpha=\beta,$ we pick $(0,0)$ and $\left(\alpha,\frac{1}{\alpha}\right)$, to see that $$\sqrt{\alpha^2+1}\leq\sqrt{\alpha^4+\alpha},$$ which is a contradiction.
For the second part note that on the line $L_x$, we have that $$d(T(x,y),T(x,z))=\sqrt{\alpha^2(y-z)^2}=\alpha d((x,y),(x,z))$$ for all $y,z\in\mathbb R$. Therefore, $T$ is a contraction on $L_x$. The fixed point of $T$ is $(x,0)$, since $(x,\alpha y)=(x,y)$, so $y=0$.
$\textbf{Where I am Stuck:}$ I cannot get a handle on the contradiction when $\alpha<\beta.$ Could someone give me a hint on this part of the problem?
Thank you for your time and appreciate any comments on my attempt at the problem, and on the part on which I am stuck. Also, if my attempt at the other parts of the problem is not clear, please leave a comment and I will clarify my solution to the best of my ability.