I think I don't see the wood for the trees: In my notes I found the remark that if $p:E \rightarrow B$ is a covering map, then for each $b \in B$ we have that $p^{-1}(b)$ in $E$ has the discrete topology. This is of course equivalent to show that every $x \in p^{-1}(b)$ is open.
Now the comment is: Each slice of the neighborhood around $B$ is open and intersects $p^{-1}(b)$ in a single point, therefore this point is open. ?!
I mean sure if I take a neighborhood $b \in U \subset B$ then $p^{-1}(U)= V_1 \cup ...\cup V_n$. Hence, $V_i \cap p^{-1}(b)$ is a single point, but how does this mean that this point must be open?-Cause we don't know that $p^{-1}(b)$ is open.
This is just from the definition of subspace topology: a set is open in the fiber iff it is the intersection with the fiber of an open set in E.