In Munkres' topology book, the following claim is made:
If $p : E \to B$ is a covering map, then for every $b \in B$, $p^{-1}(b)$ is a discrete subspace of $E$.
Here's my attempt at a proof:
Given $b \in B$, there's an open set $U \ni b$ and disjoint open sets $\{V_i\}_{i \in I}$ in $X$ such that $p^{-1}(U) = \bigcup_{i \in I} V_i$ and such that $p \big|_{V_i} : V_i \to U$ is a homeomorphism. Hence, if $x \in p^{-1}(b) \subseteq p^{-1}(U)$, then $x \in V_i$ for some $i \in I$. If $y \in p^{-1}(b) \cap V_i$, then $p(y) = b$ and $y \in V_i$. But $p \big|_{V_i}$ is, in particular, injective so $p \big|_{V_i}(x) = b = p \big|_{V_i}(y)$ implies $x=y$. This shows $p^{-1}(b) \cap V_i = \{x\}$, proving that $\{x\}$ is open in $p^{-1}(b)$.
Does this seem right?
I'm just going to put this as an answer since the OP has basically answered his/her own question perfectly.
Yes, the intuition is to imagine the $V_i$ as a stack of pancakes that get mapped down to the plate (the set $U$) by projection. Small quibble: I think $X$ should be $E$.