Let $w_1, w_2$ be two complex numbers that are linearly independent over $\mathbb{R}$, i.e., $\frac{w_1}{w_2}$ does not belong to $\mathbb{R}$.
Let $\mathbb{T}= \mathbb{Z} w_1+ \mathbb{Z}w_2$.
We define an equivalence relation on $\mathbb{C}$ by saying that two complex numbers $z_1, z_2$ are equivalent if and only if $z_1-z_2 \in \mathbb{T}$.
Let $\mathbb{C}/\mathbb{T}$ be the set of all equivalence classes of the equivalence relation defined above.
Let $\pi:\mathbb{C}\rightarrow \mathbb{C}/\mathbb{T}$ be the canonical projection.
We define a topology $\tau$ on $\mathbb{C}/\mathbb{T}$ by saying that a subset $U$ of $\mathbb{C}/\mathbb{T}$ is open if and only if $\pi^{-1}(U)$ is an open subset of $\mathbb{C}$.
How to prove that $\mathbb{C}/\mathbb{T}$ with the topology $\tau$ defined above is compact?
Consider the subset $$D = \{a w_1 + b w_2 \mid (a,b) \in [0,1] \times [0,1]\} \subset \mathbb C $$ Clearly $D$ is homeomorphic to $[0,1] \times [0,1]$ and so is compact. Also, the restriction of $\pi : \mathbb C \to \mathbb C / \mathbb T$ to the subset $D$ is surjective, because every equivalence class contains an element of $D$. To see why this is true, start with an arbitrary element of $\mathbb C$, written in the form $x w_1 + y w_2 \in \mathbb C$ (where $x,y \in \mathbb R$). Take $a = x - \lfloor x \rfloor$ and $b = y - \lfloor y \rfloor$, where $\lfloor t \rfloor$ is the largest integer $\le t$. It follows that $a w_1 + b w_2 \in D$, and that $a w_1 + b w_2$ is equivalent to $x w_1 + y w_2$, because $$(x w_1 + y w_2) - (a w_1 + b w_2) = \lfloor x \rfloor w_1 + \lfloor y \rfloor w_2 \in \mathbb Z w_1 + \mathbb Z w_2 $$ Therefore, $\mathbb C / \mathbb T$ is the continuous image of the compact set $D$, hence is compact.