Why does the non-euclidean distance between the lines $x=0$, $x=1$ approach $0$ as $y \to \infty$?
Please see http://books.google.ca/books?isbn=0387290524 on pg 191 for more information.
My thoughts: find the noneuclidean line that contains both points $iy$ and $iy+1$. Then transform to $y$-axis by Möbius transformation $f$. Then as $y$ goes to $\infty$, $f(iy)=f(iy+1) = i$.
I just don't know how to calculate the limit of $f(iy)$ as $y\to\infty$.
Your question body and the book itself say $y \to \infty$, so I'll assume that the title is wrong.
Your idea is a good one. We want to find the limit of $d(iy, iy + 1)$ as $y \to \infty$. We know that the non-Euclidean distance is invariant under dilations. Apply the dilation $z \mapsto \dfrac{1}{y} z$ to get:
$$ d(iy, iy + 1) = d\left(i, i + \frac{1}{y}\right) $$
The distance between $i$ and $i + 1/y$ goes to $0$ as $y \to \infty$. The desired result follows.