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.

$\mathbb{R}^2 \setminus \mathbb{Z}^2$ is a union of translated strips that look like $(0, 1) \times \mathbb R$ and $\mathbb R \times (0, 1)$, each of which is open.
