Let $p:E \rightarrow X$ be a covering space.
It is well known that if $X$ is connected, then all the fibers have the same cardinality. This can be seen as a simple consequence of the fact that the cardinality of $p^{-1}(x)$ is a locally constant map.
My question is, assuming $E$ is path-connected, how can I prove that all the fibers have the same cardinality? Can somebody help me?
Thanks.
-- EDIT --
Idea : If $E$ is path connected, then $X$ is connected and we're done. To see that $X$ is connected, let $x,y\in X$ and let $x_{0}\in p^{-1}(x)$ and $y_{0}\in p^{-1}(y)$. Since $E$ is path connected, let $\sigma$ be a path connecting $x_{0}$ to $y_{0}$. Then, $p\circ\sigma$ is a path connecting $x$ to $y$. Thus, $X$ is path connected and hence, connected.
another way you can say this by using uniqueness of path-lifting
given two points $x$ & $y \in X$ you can consider a path from $x$ to $y$ (this will exists since $E$ is path-connected and $p$ is continuous)...then given any point $x_0 \in p^{-1} (x)$ you can get a lift of that path starting from $x_0$ ...and by the uniqueness there will be a unique $y_0$ where the lift of the initial path will end..this gives a injective map from the set $p^{-1}(x) \to P^{-1} (y) $ ...and by using same argument we'll get a injective map $p^{-1}(y) \to P^{-1} (x) $ ...this will prove that all the fibers have same cardinality.
(but honestly speaking some how in this proof also I am using the fact that $p$ is locally homeomorphism...so this proof is not very far away from the proof you've mentioned)