Is this statement true?(covering map) 
Let $C,X$ be topological spaces.
Let $p:C\rightarrow X$ be a continuous function.
Let $U$ be an evenly covered open subset of $X$.
Let $V$ be an open subset of $C$ such that $p|_V:V\rightarrow U$ is a homeomorphism.
Then, there exists a mutually disjoint family $\{V_i\}$ of open subsets of $C$ such that $p|_{V_i}:V_i\rightarrow U$ is a homeomorphism and $V=V_i$ for some $i$ and $p^{-1}(U)=\bigcup V_i$.

I have proved this on my own to figure out what the definition of a sheet would be and I want to verify this.
Related: What is the definition of "sheet"?
 A: An open set $U \subseteq X$ is said to be evenly covered if there is a collection of disjoint open sets $\{U_i\}_{i\in I}$ in $C$ such that $p^{-1}(U) = \bigcup_i U_i$ and $p|_{U_i} : U_i \to U$ is a homeomorphism. We say that $\{U_i\}$ evenly cover $U$.
Suppose for now that $U$ is connected, and hence so are the sets $U_i$. If $V \subseteq C$ is an open set such that $p|_V : V \to U$ is a homeomorphism, then $V$ is connected and $V \subseteq p^{-1}(U)$. As $p^{-1}(U)$ is the disjoint union of the sets $U_i$, all of which are connected, they are the connected components of $p^{-1}(U)$ and hence $V \subseteq U_i$ for some $i$. For any $x \in U_i$, $p(x) \in U$ and as $p|_V : V \to U$ is surjective (it is a homeomorphism), there is $y \in V$ such that $p(y) = p(x)$. But $p|_{U_i} : U_i \to U$ is a homeomorphism, so $y = x$ and hence $V = U_i$. 
That is, if $U$ is evenly covered by $\{U_i\}$ and $V \subseteq C$ is an open set such that $p|_V : V \to U$ is a homeomorphism, then $V = U_i$ for some $i$. If $U$ is not connected, then this need not be true. For example, consider $p : S^1 \to S^1$ given by $p(z) = z^2$. The set 
$$U = \{(\cos\theta, \sin\theta) \mid \theta \in (0, \pi)\cup (\pi, 2\pi)\}$$ 
is evenly covered by $U_1$ and $U_2$ where
\begin{align*}
U_1 &= \left\{(\cos\theta, \sin\theta) \mid \theta \in \left(0, \frac{\pi}{2}\right)\cup\left(\frac{\pi}{2}, \pi\right)\right\}\\
U_2 &= \left\{(\cos\theta, \sin\theta) \mid \theta \in \left(\pi, \frac{3\pi}{2}\right)\cup\left(\frac{3\pi}{2}, 2\pi\right)\right\}.
\end{align*}
However, the set
$$V = \left\{(\cos\theta, \sin\theta) \mid \theta \in \left(0, \frac{\pi}{2}\right)\cup\left(\frac{3\pi}{2}, 2\pi\right)\right\}$$
is an open subset of $S^1$ such that $p|_V : V \to U$ is a homeomorphism, but $V \neq U_1, U_2$. However, unlike in the connected case, $U$ is evenly covered in more that one way. In this case, $U$ is also evenly covered by $V_1$ and $V_2$ where
\begin{align*}
V_1 &= \left\{(\cos\theta, \sin\theta) \mid \theta \in \left(0, \frac{\pi}{2}\right)\cup\left(\frac{3\pi}{2}, 2\pi\right)\right\}\\
V_2 &= \left\{(\cos\theta, \sin\theta) \mid \theta \in \left(\pi, \frac{3\pi}{2}\right)\cup\left(\frac{\pi}{2}, \pi\right)\right\}
\end{align*}
and $V = V_1$. This is indicative of the general situation. That is, if $U$ is openly covered by $\{U_i\}$, and $V \subseteq C$ is an open subset such that $p|_V : V \to U$ is a homeomorphism, then $V$ need not be equal to $U_i$ for any $i$, but $U$ is evenly covered by $\{V_i\}$ where $V = V_i$ for some $i$.
Let $U = \bigcup_{\alpha \in A} U_{\alpha}$ where $\{U_{\alpha}\}$ are the connected components of $U$. Likewise, let $V = \bigcup_{\alpha} V_{\alpha}$ and $U_i = \bigcup_{\alpha} U_{i,\alpha}$ where $V_{\alpha}$ and $U_{i,\alpha}$ are the connected components of $V$ and $U_i$ respectively, and $p(V_{\alpha}) = p(U_{i,\alpha}) = U_{\alpha}$. Note that $\{U_{i,\alpha}\}_i$ evenly covers $U_{\alpha}$.
Let $I_0 = \{i \in I \mid U_i\cap V \neq \emptyset\}$. For $i \in I_0$, note that $U_i\cap V = \bigcup_{\alpha \in A_i}U_{i,\alpha}$ where $A_i \subseteq A$; note that $\{A_i\}_{i\in I_0}$ is a partition of $A$. For $i \not\in I_0$, let $V_i = U_i$. Now fix $j \in I_0$ and set $V_j = V$. For $i \in I_0\setminus\{j\}$, set $V_i = \bigcup_{\alpha \in A_i}U_{j,\alpha}\cup\bigcup_{\alpha\in A\setminus A_i} U_{i,\alpha}$. Then $\{V_i\}_{i\in I}$ evenly covers $U$ and $V = V_j$.
