Here's an argument for why you can't have any elements $\beta \in I\setminus S$ such that $\beta$ is in the square with vertices $\{0, \alpha, i\alpha, (1+i)\alpha\}$. Let's suppose, for a contradiction, that you had such a $\beta$. Let $d_{1}$ be the distance between $\beta$ and $0$, and let $d_{2}$ be the distance between $\beta$ and $(1+i)\alpha$. I claim that $$\min\{d_{1}, d_{2}\} <\sqrt{N(\alpha)}.$$
What is $\sqrt{N(\alpha)}?$ It is the distance from $\alpha$ to $0$, hence it is the side length of the square. The points that are distance $\sqrt{N(\alpha)}$ or less from $0$ lie in a quarter-circle of radius $\sqrt{N(\alpha)}$ centered at $0$, and similarly for $(1+i)\alpha$. If you draw the picture, you see that the union of these two quarter-circles covers the entire square, hence every point in the square has either distance less than or equal to $\sqrt{N(\alpha)}$ from $0$, and/or distance less than or equal to $\sqrt{N(\alpha)}$ from $(1+i)\alpha.$ Furthermore, note that equality is only achieved at the vertices $\alpha, i\alpha.$
Since our element $\beta$ is not in $S$, it is not equal to $\alpha, i\alpha,$ so the inequalities are strict, and we have $\min\{d_{1}, d_{2}\} <\sqrt{N(\alpha)}$ as claimed. If $d_{1} < \sqrt{N(\alpha)}$, then $d_{1}^{2} < N(\alpha).$ But $d_{1}^2$ is the square of the distance from $\beta$ to $0$, so $d_{1}^{2} = N(\beta).$ This contradicts the minimality of $N(\alpha).$
On the other hand, suppose $d_{2} < \sqrt{N(\alpha)}.$ Note that the distance $d_{2}$ between $b$ and $(1+i)\alpha$ is the same as the distance between $\beta - (1+i)\alpha$ and $0.$ So, by the same argument, the norm of $\beta - (1+i)\alpha$ is less than the norm of $\alpha$. But $\beta - (1+i)\alpha$ is an element of the ideal, contradiction.
So we've shown that you don't have any points of $I$ in the square $\{0, \alpha, i\alpha, (1+i)\alpha\}$ which aren't the vertices. Using the translation idea in the previous paragraph, can you do this for all the other squares?
