Let $q:X \rightarrow Y$ and $r:Y \rightarrow Z$ be covering maps and $p= r \circ q$. If $r^{-1}(z)$ is finite for all $z \in Z$, then $p$ is a covering map.
Now I found the following proof:
Let $z \in U_z$ where $U_z$ is some open set that is evenly covered by $r$.
Let $\{V_1,...,V_k\}$ be a partition of slices with $y_i \in V_i$ with $r(y_i)=z$. Then there exists for all $y_i$ an open set $W_i$ that is evenly covered by $q$ and without loss of generality $W_i \subset V_i$. Set $U :=r(W_1) \cap...\cap r(W_k)$ which is open in $Z$, due to the fact that covering maps are open, then $U$ is evenly covered by $p$.
I don't get it why $U$ works. I mean, it is not that clear to me why $r^{-1}(U) \subset W_i$? But we should have $r^{-1}(U) \subset W_i$ in order to be sure that $r^{-1}(U)$ is evenly covered by $q$.