If $\inf_{k\in\mathbb{Z}}|x-y+k|\leq\frac{1}{2N}$, then $\inf_{k\in\mathbb{Z}}|Nx-Ny+k| = N\inf_{k\in\mathbb{Z}}|x-y+k|$.

Question

Suppose that $x,y\in [0,1)$ and define $$d(x,y) = \inf_{k\in\mathbb{Z}}|x-y+k|.$$ Let $N\in\mathbb{N}$ and assume that $d(x,y)<\frac{1}{2N}$, then I want to prove that $$d(Nx,Ny) = Nd(x,y),$$ without refering to $S^{1}$. The inequality $d(Nx,Ny)\leqslant Nd(x,y)$ is clear, even without the assumption that $d(x,y)<\frac{1}{2N}$. I really struggle with the formalities of the other inequality. Any help would be greatly appreciated!

EDIT: I can prove that $$|Nx-Ny+k|+|k|\geq Nd(x,y)$$ for all $k\in\mathbb{Z}$. But again without using the asumption that $d(x,y)<\frac{1}{2N}$.