Let $K/\mathbb{Q}$ be a real quadratic number field, let $\sigma_1,\sigma_2 \colon K \to \mathbb{R}$ be the two real embeddings and let $\lambda \colon K \to \mathbb{R}^2$ be defined as $x\mapsto (\sigma_1(x),\sigma_2(x))$. We know from Dirichlet's Unit Theorem that $$ \mathcal{O}_{K}^{\times } = \pm \varepsilon^{\mathbb{Z}} $$ for some $\varepsilon \in \mathcal{O}_{K}^{\times }$, which in fact can be uniquely determined if we impose $\sigma_1(\varepsilon )>1$.
Let $J$ be a fractional ideal in $K$. Then $\lambda(J)\subseteq \mathbb {R}^2$ is a lattice and we have an action of $\mathcal{O}_{K}^{\times }$ on this lattice defined as $u \cdot \lambda(x)=\lambda(ux)$. Since $\sigma_1(ux)\sigma_2(ux)=N_{K/\mathbb{Q}}(ux)=\pm N_{K/\mathbb{Q}}(x)=\sigma_1(x)\sigma_2(x)$, the orbits of this action look like discrete hyperbolic lines.
Now consider the region $A=\{(x,y)\in\mathbb{R}^2 \mid 1 < \frac{x}{y} \leqslant \frac{\sigma_1(\varepsilon)}{|\sigma_2(\varepsilon)|} = \sigma_1(\varepsilon)^2 \}$, which is a region of the plane between two lines passing through the origin contained in the first and third quadratns.
So the claim now is that for any $\lambda(x)\neq (0,0)$ in this lattice there is a unique integer $n\in \mathbb{Z}$ such that $$ \varepsilon^n \cdot \lambda(x)\in A$$
I have the feeling that this is a very elementary statement and that it is just some inequality that I am not seeing. If we let $\varepsilon^n$ act on $\lambda(x)$, then we get (say for example that $\sigma_2(\varepsilon)>0$, although I am not sure if it will always be $<0$ as in $\varepsilon = 1+\sqrt2$): $$ \frac{\sigma_1(\varepsilon^n x)}{\sigma_2(\varepsilon^n x)}=\sigma_1(\varepsilon )^{2n}\frac{\sigma_1(x)}{\sigma_2(x)}$$
So why is the claim true? Again: I suspect that this is just a very silly thing that I am not seeing. But I am really stuck, so I would appreciate any help.