In Munkres' Topology section 53, he says
If $p:E\to B$ is a covering map, then for each $b\in B$ the subspace $p^{-1}(b)$ of $E$ has the discrete topology. For each slice $V_\alpha $ is open in $E$ and intersects the set $p^{-1}(b)$ in a single point; therefore, this point is open in $p^{-1}(b)$.
I don't understand what this means. Could anyone explain it with some example? Especially, I want to understand why for each $b\in B$ the subspace $p^{-1}(b)$ of $E$ has the discrete topology. And how each slice $V_\alpha $ intersects the set $p^{-1}(b)$ in a single point.
He also gives an exercise which I think is related to the remark.
Let $p: E\to B$ be a covering map; let $B$ be connected. Show that if $p^{-1}(b_0)$ has $k$ elements for some $b_0 \in B$, then $p^{-1}(b)$ has $k$ elements for every $b \in B$.
By construction, $p_{\vert V_\alpha}$ is a homeomorphism onto an open neighborhood of $b$ in $B$, so that the point $b$ has exactly one antecedent by $p$ in $V_{\alpha}$.
By definition of the induced topology, $p^{-1}(b)\cap V_{\alpha}$ is an open set of $p^{-1}(b)$, since $V_\alpha$ is open in $E$.
Whence the answer to your first question.
For the second one, notice that $B$ is the disjoint union of the sets $B_k$ formed of $b\in B$ such that $p^{-1}(b)$ has cardinality $k$, for $k\in\mathbb{N}\cup\{+\infty\}$ and that each $B_k$ is open. Indeed, if $b\in B_k$, since $p$ is a covering map.