Consider the covering space $\widetilde X= \mathbb R$ of $X=S^1$ with $p:\mathbb R \to S^1,r\mapsto(\cos(2\pi r),\sin(2\pi r))$ as the covering map. The fibre of $(1,0)$ is $\mathbb Z$. Let $U$ be the open disconnected neighborhood of $(1,0)$ which is the disjoint union of two open arcs $U_1=\{(x_1,x_2)\in S^1\mid x_1>0\}$ and $U_2=\{(x_1,x_2)\in S^1 \mid x_1<0\}$. Then the preimage is the disjoint union of open intervals of length $\frac12$ centered at the points of $\frac12\mathbb Z$, say $J_n=(\frac n2-\frac14\ ,\ \frac n2+\frac14)$. For an open set $V_\alpha$ which is supposed to be mapped to $U$, you can take $V_\alpha=J_{2n}\cup J_{2n+1}$ but you could also take $V_\alpha=J_{2n}\cup J_{2n-1}$ or even $V_\alpha=J_{2n}\cup J_{2n-k}$ for an odd integer $k$.
On the other hand, if $U$ is connected, then the partition of its preimage is unique, since the $V_\alpha$ are the connected components.
Let $p: E \rightarrow B$ be a covering map. For a fixed finite number $k$, define $A_k = \{ b \in B: |p^{-1}(b)|=k \}$. I claim this set is open and closed in $B$.
To see it is open, let $b \in A_k$ and let $U$ be an evenly covered neighbourhood of $b$. So $p^{-1}[U]= \cup_{i \in I} V_i$, where the $V_i$ are pairwise disjoint, and for every $i$, $p$ restricted to $V_i$ is a homeomorphism between $V_i$ and $U$. Then, as all these restrictions are bijections in particular, $b$ has one pre-image in each $V_i$, so $|I|=k$, and moreover, we know that $|p^{-1}(y)|=k$ as well for every $y \in U$, so that $p \in U \subset A_k$, making $b$ an interior point of $A_k$.
Suppose now that $b \notin A_k$, and we take an evenly covered neighbourhood $U$ of it, with the same notation as before. The same reasoning as the previous paragraph gives that all points in $U$ have the same number of preimages as $b$ has, so if $b \notin A_k$, we also have $U \subset B \setminus A_k$, which shows that all points not in in $A_k$ are also not in its closure, so $A_k$ is closed as well.
Now, as $B$ is connected, the only sets that can be open and closed are $B$ and $\emptyset$, so assuming there is some point with finite fibre, all other points have the same sized fibre as well.
Best Answer
Let $\{V_\alpha\}$, $\{W_\beta\}$ be partitions of $p^{-1}(U)$.
Pick $\alpha$ and consider $V_\alpha$, choose $e \in V_\alpha$. Since $e \in p^{-1}(U) = \bigsqcup_\beta W_\beta$, there exists a unique $\beta$ such that $e \in W_\beta$.
We want to show that $V_\alpha = W_\beta$. Suppose by way of contradiction that $V_\alpha \neq W_\beta$. Then $V_\alpha \setminus W_\beta \neq \emptyset$. We know $V_\alpha \cap W_\beta \neq \emptyset$ since each contain $e$. Notice that $V_\alpha \cap W_\beta$ is open since it is the intersection of two open sets. Moreover $V_\alpha = \big( V_\alpha \cap W_\beta \big) \cup \big( V_\alpha \setminus W_\beta \big)$. And of course $\big( V_\alpha \cap W_\beta \big) \cap \big( V_\alpha \setminus W_\beta \big) = \emptyset$. If we can show that $V_\alpha \setminus W_\beta$ is open, then we have a separation of $V_\alpha$. This is a contradiction to our hypothesis that $U$ is connected since $V_\alpha$ is homeomorphic to $U$ and is this connected as well. Hence, we can conclude that $V_\alpha = W_\beta$ and thus the partition is unique.
Now why is $V_\alpha \setminus W_\beta$ open? If $e' \in V_\alpha \setminus W_\beta$, there exists $\beta_{e'} \neq \beta$ such that $e' \in W_{\beta_{e'}}$. Notice that $V_\alpha \setminus W_\beta = V_\alpha \cap \bigcup_{e' \in V_\alpha \setminus W_\beta} W_{\beta_{e'}}$ which is open because $\bigcup_{e' \in V_\alpha \setminus W_\beta} W_{\beta_{e'}}$ is open since it is the union of open sets, and the intersection of two open sets is open.