Let $p:\tilde X\to X$ be a covering. Define an equivalence class on $X$ as follows $$x\sim z \Leftrightarrow |p^{-1}(x)|=|p^{-1}(z)| $$ where $||$ means the cardinality of the fiber. Take any $x\in X$ I want to show that the equivalence class $[x]$ is open in $X$.
My try: I want to show that $[x]$ is a neighborhood of all its points. So take $y\in [x]$ and show that $[x]$ is a neighborhood of $y$; that is $[x]$ contains an open set that contains $y$.
Since $p$ is a covering, there exists open neighborhoods $V_x$ of $x$ and $V_y$ of $y$ such that there exist a homeomorphisms $$g:p^{-1}(V_x)\to V_x\times I\;\;\text{and}\;\;h:p^{-1}(V_y)\to V_y\times J$$ where $I$ and $J$ are discrete sets.
The restrictions of $g$ and $h$ to $p^{-1}(x)$ and $p^{-1}(y)$ give the following homeomorphisms $$g:p^{-1}(x)\to \{x\}\times I\;\;\text{and}\;\;h:p^{-1}(y)\to \{y\}\times J$$
Now since $y\in [x]$, then there exists a homeomorphism $s:\{x\}\times I\to \{y\}\times J$ which is possible only if $I=J$. Now i'm tempted to say that $[x]$ contains the open set $V_y$ that contains $y$ which finishes the proof but i don't see how to conclude. Thanks for your help!
I think you are really almost done. For every point $y'$ in $V_y$, you can look at its fiber, which will be isomorphic to $\{y'\}\times I$ and hence has the same cardinality as the fiber of $x$. So all the points of $V_y$ lie in $[x]$.