Prove that if $p:E \rightarrow X$ is a covering map with $E$ connected and $p^{-1}(x_{0})$ has $k$ elements for some $x_{0} \in X$, then $p^{-1}(x)$ has $k$ elements for every $x \in X$.
Is my proof correct?
Let $A=\{x \in X : |p^{-1}(x)|=k\}$ and $B= \{x \in X : |p^{-1}(x)|\neq k\}$; then $X = A \cup B$. For each $x \in X$, let $U_{x}$ be a neighborhood which is evenly covered by $p$. Then, since the number of sheets over $U_{x}$ is equal to $|p^{-1}(u)|$ for any $u \in U_{x}$, it follows that $U_{x} \cap U_{y} = \emptyset$ if $x \in A$ and $y \in B$. Now, let $C= \cup_{x\in A}U_{x}$ and $D=\cup_{x \in B}U_{x}$. We have $C$ and $D$ are disjoint open sets with $X=C \cup D$; thus $D$ must be empty or this would be a separation of $E$ contradicting connectedness.