is $\{(x,y) : x,y \in \mathbb{Z} \}$ a closed set? I claim yes, and to show this, it will suffice to show that $\mathbb{R}^2 \setminus \mathbb{Z}^2$ is open. So that for every $x \in \mathbb{R}^2 \setminus \mathbb{Z}^2$, we must find a neighborhood $N$ of $x$ such that $N \cap \mathbb{Z}^2 = \varnothing$. Let $r = \min(\|x\|, 1 - \|x\|)$ in $N = D^2(x,r)$. Suppose there exists $z \in N \cap \mathbb{Z}^2$. So
$$ \|z\| \leq \|z-x\| + \|x\| < \min(\|x\|, 1 - \|x\|) + \|x\| \leq 1 - \|x\| + \|x\| = 1 $$
So, we have a contradiction, and therefore $N \cap \mathbb{Z}^2$ must be empty as desired.
Is this correct? any feedback? thanks.

I think your idea with the norm is not exactly the good one. Take $(x,y) \in \mathbb R^2 \backslash \mathbb Z^2$. Then either $x$ or $y$ is not an integer. Without loss of generality, suppose $x$ is not an integer. Then there exists $z_1, z_2 \in \mathbb Z$ such that $z_1 < x < z_2$. The set $]z_1,z_2[ \, \times \, \mathbb R$ contains $(x,y)$ and is open, so it is a neighborhood of $(x,y)$. It contains no point in $\mathbb Z^2$, so is a subset of $\mathbb R^2 \backslash \mathbb Z^2$, and thus $\mathbb Z^2$ is closed. 
Feedback : I think you tried to compute an open ball whose radius is smaller than something so that your ball doesn't intersect $\mathbb Z^2$, but it's just not working out (at least the way you wrote it). I took an entire open strip of the plane that doesn't cross $\mathbb Z^2$ : you could extract an open ball from it if you wanted (for instance, it could be centered at $(x,y)$ and have radius $\min \{ |x - z_1|, |x-z_2| \}$).
Hope that helps, 
