Let $X:= \mathbb{R_1} \sqcup \mathbb{R_2}$ where $\sqcup$ means the disjoint union. Now $X$ is a topological space with the "Disjoint topology", in which the opens are disjoint union of opens of $\mathbb{R_1}$ and of $\mathbb{R_2}$.
I have the projection $\pi:X \rightarrow \mathbb{R}$ and I can take the induced topology on $\mathbb{R}$ (the opens are the ones with pre-image an open of $X$).
Now $\pi$ is not a covering for $\mathbb{R} $.
I don't know precisely how to show this, my intuition is that if I can take an open interval $I \in \mathbb{R}$, then $\pi^{-1}(I)$ will be connected in $X$ but not arc connected and so there is no omeomorphism between the "components" of the pre-image of $I$ and $I$.
Am I right?