Quick way to show $\mathbb{R}^2/ \mathbb{Z}^2$ is compact See title - I can think of 2 ways:
1) show it's homeomorphic to a torus, and apply Heine-Borel in $\mathbb{R}^3$. The problem here is  showing its a homeomorphism is fiddly
2) show that $\mathbb{R}^2/ \mathbb{Z}^2 \cong [0,1]\times[0,1]/$~ where ~ is the usual equivalence relation, then apply products and quotients of compact spaces are compact.
Is there a better way?
 A: Let $i : \mathbb{R^2} \to \mathbb{R^2} /\mathbb{Z^2}$ be the quotient map. If $U_\alpha$ is an open cover of $\mathbb{R^2} /\mathbb{Z^2}$, then $i^{-1}(U_\alpha)$ is an open cover of $\mathbb{R^2} $, in particular, it covers the compact set $[0,1]^2$ and hence has a finite subcover.
A: If you know that $\Bbb R/\Bbb Z$ is compact (which is the same kind of problem but lets assume you know that already), and if you want to use lots of little arguments of topology you could do as follows. The map $f:\Bbb R^2\to \Bbb R/\Bbb Z\times \Bbb R/\Bbb Z$ defined by 
$$f:(x,y)\mapsto ([x],[y])$$
factors to a continuous bijection $\tilde{f}:\Bbb R^2/\Bbb Z^2\to \Bbb R/\Bbb Z\times \Bbb R/\Bbb Z$ wich is a homeomorphism if and only if $f$ is a quotient map. But $f$ is the product of two open maps ($\Bbb R\to\Bbb R/\Bbb Z$ is an open map because its the quotient by a subgroup in a topological group), so it is open hence quotient. You conclude using the fact that $\Bbb R/\Bbb Z\times \Bbb R/\Bbb Z$ is compact.
