Here are two "general" explanations for why $\pi: \mathbb{C} \rightarrow \mathbb{C}/\Lambda$ is an open map.
1) $\pi$ is a covering map, hence a local homeomorphism, hence an open map.
I think this is quite intuitive and easy to see: about any point on the torus $\mathbb{C}/\Lambda$, the preimage of a sufficiently small disk-shaped neighborhood will be a "lattice" of small disks in $\mathbb{C}$. In fact $\pi$ is a regular covering map, hence the quotient by an action of the group $\Lambda$.
This leads to a more general answer.
2) For a group $G$ acting on a topological space $X$, the quotient map $\pi: X \rightarrow X/G$ is open.
Proof: By definition of the quotient topology on $X/G$, we must show that if $U \subset X$ is open, then $\pi^{-1} \pi U$ is open. But $\pi^{-1} \pi U = \bigcup_{g \in G} g U$. Since each $g \bullet$ is a homeomorphism (that is part of the definition of a group action on a topological space) and $U$ is open, $g U$ is open, so $\pi^{-1} \pi U$ is a union of open sets, hence open.
Note that Paul Garrett remarks that most reasonable quotient maps are open. While I certainly agree with this as a principle, in any particular reasonable situation one still needs to summon a proof. Studying in general the problem of openness of quotient maps seems (to me, of course) to be unrewarding and technical -- c.f. Bourbaki's General Topology, which does entirely too much of this for my taste -- so it is worthwhile to collect "easy" explanations like those above. In fact 2) above was taken directly from lecture notes for a course on modular curves I taught recently: and it is from page 1 of those notes!