I'm reading from "Galois Groups and Fundamental Groups" of Szamuely about covering spaces.
Given an even, a.k.a. properly discontinuous, continuous action of a group $G$ on $Y$ topological space. Then there is the following:
Lemma: If $Y$ is connected, then the projection $\pi_G : Y \rightarrow Y/G $ is a covering map of $Y$ onto $Y/G$.
Proof: The map is surjective, then given $x \in Y/G$ there exists an open neighbourhood $V \ni x$ such that $V= \pi_G (U)$ and $U$ has the property that $\{ gU \}_{g \in G}$ are pairwise disjoint. So $\pi_G^{-1}(V)= \coprod_\limits{g \in G} gU $, from which the result follows.
Now, my question is: where is the need of connectedness of $Y$? I mean, given a non connected $Y = \coprod Y_\alpha$, where $Y_\alpha$ are the connected components, then because of the action of the group is even, I can repeat the same argument of the proof above and notice that the fiber of $V$ will be distributed all over the connected components, but the fundamental properties of a covering, i.e. fibre which is a disjoint union of opens, everyone homeomorphic to $V$, depends only on the action of the group .
The behavior of the action can change in different connected components, so in general you have no way to guarantee that the fibers of points in distinct components are homeomorphic.
As a simple example consider $Y=\{a,b,c\}$ with the discrete topology and a two element group $G$, where the non-trivial element swaps $a$ and $b$. Then $Y/G$ has two points, and the fibers are the non-homeomorphic sets $\{a,b\}$ and $\{c\}$.
Edit: If in the definition of covering you don't require the fibers to be homeomorphic, your argument does indeed work, i.e. the lemma is valid also for a disconnected $Y$.
Note that some authors may require connectedness of the base space in the definition of covering, since some properties will be only locally constant over the base space (e.g. the degree, as shown in the above example).