While reading Hatcher, he stated "If $p: \tilde{X} \rightarrow X$ is a covering space, then the cardinality of the set $p^{-1}(x)$ is locally constant. I have trouble seeing that this is the case.
I suppose we would want to take a neighborhood $U$ of $x$ then take another point $y \in U$. We would want to look at the path from $x$ to $y$ and this should lift to a path in $\tilde{X}$. But I fail to see how to proceed to show that the cardinalities of $p^{-1}(x)$ and $p^{-1}(y)$ should be the same. Moreover, how would we choose our $U$ 'sufficiently small' for this to be so? Is it the case that if $X$ is connected then all of the fibers have the same cardinality?
In the definition of a covering space you have a covering of $X$ by open sets $U_i$ such that $p^{-1}(U_i)\cong U_i\times F$. Then the cardinality of $p^{-1}(x)$ is $|F|$ for any $x\in U_i$.