$X,Y$ are path connected and Hausdorff, $p:X\rightarrow Y$ is a cover map, $y\in Y$ and $|p^{-1}(y)|<\infty$ then $|p^{-1}(y')|=|p^{-1}(y)|$ for all $y'\in Y$
In case $Y$ is compact, I figured out the next proof idea, but I don't see how to make it in case $Y$ is not compact.
Compact case:
Let $p^-1(y)= \{x_1,x_2,...x_n\}$ and $y \in V_y$ the elementary set such that $p^{-1}(V_y) = \biguplus_1^nU_i$. So from $U_i$ disjointedness $\forall i\in [n]: x_i\in U_i$. For every $y'\in Y$ denoting $q:[0,1]\rightarrow Y$ a path between $y$ and $y'$ one may take an open cover of $q$ by elementary sets $V_{q(s):s\in[0,1]}$, take a finite sub-cover and then work with the fact that $p|_{U_i}:U_i \rightarrow V_y$ is an homeomorphism and from it to conclude the that for every $y=q(s)$, $|p^{-1}(y)| = n$.
Compactness is irrelevant for this problem (and it's not a given).
We only need connectedness (no paths) for $Y$:
Let $y_0 \in Y$. Then by assumption $p^{-1}[\{y_0\}]$ is a finite subset of $X$, say of size $n_0 >0$. Define $A = \{ y \in Y: |p^{-1}[\{y\}]| = n_0 \}$, we know it's non-empty as $y_0 \in A$. If $y \in A$ and $V_y$ is the evenly covered neighbourhood of $y$ for $p$, for every $y' \in V_y$ we have that $|p^{-1}[\{y'\}]|=n_0$ as well (check this; $p^{-1}[V_y]$ must have $n_0$ disjoint open sets, each of which maps onto $V_y$, so all these "composants" contain a single pre-image for $y'$ as well). So $y \in V_y \subset A$, showing that $y$ is an interior point of $A$. So $A$ is open.
A very similar argument shows that $Y\setminus A$ is open as well (also use the evenly covered neighbourhood). So $A$ is closed-andopen non-empty so equals $Y$ as $Y$ is connected. Done.