Geometric Proof that $\mathbb{Z}[\sqrt{-3}]$ is non-Euclidean 
Is there a geometric proof showing that $\mathbb{Z}[\sqrt{-3}]$ is non-Euclidean? 

I think this is a sketch of how to proceed. Consider the elliptical region $x^2+3y^2<1$. We can then partition the plane into unit squares -- four of which contain parts of our region. Now, can we translate these pieces on top of a single unit square and examine what points are not covered by some new translated part of the ellipse?
Using this, can we find elements of $\mathbb{Z}[\sqrt{-3}]$ such that the division algorithm is invalid? I'm not really sure how to approach this; any help is greatly appreciated.

Arrange that the center of the ellipse lies on the center of one of the squares. The ellipse passes precisely through the four vertices of the square. This implies that your ring is almost but not quite Euclidean, at least with respect to the usual norm. But to complete your proof, you need to prove it is non Euclidean with respect to any multiplicative norm. In this case, it is easier to show that your ring is not a UFD. You only need a single counter example.
