Choose $\Delta=\frac{1}{2}$ (I believe this value should work). let $\delta > 0$ and let $\textbf{x}_1=(x_1,y_1) \in X$. I assuming that $d$ is the Euclidean distance.
Somehow I think we need to pick two points with the same $x$ value$(=x_1)$ that are somewhat close enough in the vertical direction. Having problems writing this formally.
I start by writing $T^n(x_1,y_1)=(x_1+ny_1 \mod 1, ny_1 \mod 1)$ and $T^n(x_1,y_2)=(x_1+ny_2 \mod1, ny_2 \mod 1)$. Then I cannot see how to proceed. I cannot see how to choose my $n$ or my $\textbf{x}_2=(x_1,y_2) \in X$ such that the $y$'s are right distance away.
Apparently a hint is that somehow $x_1+ny_1=x_1+ny_2 +n(y_2-y_1)$.
Choose $\Delta\leq1/2$, for a point $(x_1,y_1)$ and $\delta>0$ we choose $m$ such that $1/2^{m+1}<\delta$ and let $x_2=x_1$ and $y_2=y_1+1/2^{m+1}$. Let $n=2^m$ then we have that $n(y_2-y_1)=1/2$ and so $T^n(x_1,y_1)=T^n(x_2,y_2)+(1/2,0)$ and we see the two points have distance 1/2 greater than or equal to $\Delta$.