Suppose that $p: X \to B$ is a covering space where $X$ its not necessarily arc-connected. Prove that $p|_{X_0} :X_0 \to B$ is a covering space. Let $B$ be path connected and locally path connected. Suppose that $p \colon X \to B$ is a covering space where $X$ is not necessarily path connected. Let $X_0$ be a path component of $X$. Prove that $p|_{X_0} \colon X_0 \to B$ is a covering space.
For this I have to show 3 things. That:


*

*$p|_{X_0} : X_0 \to B$ is a surjective and continuous function (which can be straight forward since $p$ is a covering space).

*There exist an open set $U$ such that $p^{-1}(U)= \sqcup V_{\alpha} $ with $V_{\alpha} \subset X_0$.

*$p|_{V_{\alpha}} \colon V_{\alpha} \to U$ is a homeomorphism.
Since $p$ is a covering space there already exist a $W$ open set that satisfies 2 and 3 ($W$ is evenly covered).
Can someone help me to structure well the proof.
 A: *

*$p \mid_{X_0}:X_0 \to B$ is certainly continuous, but needn't be surjective. Luckily some definitions (for instance, Hatcher's) do not require this. 

*I think you mean that for every $b \in B$, there is some open set $U \ni b$ such that $p\mid_{X_0}^{-1}(U)$ is a disjoint union of sets in $X$ each of which is mapped homeomorphically to $U$ by $p\mid_{X_0}$ (I'm combining your 2 and 3).
We already know that such a $U$ exists because $p$ is a covering. By local path connectedness, we can assume that $U$ is path connected (otherwise replace it with a path connected set contained in $U$). Because $p$ is a cover, $p\mid_{X_0}^{-1}(U)$ is a disjoint of sets (as above) intersected with $X_0$. 
Because each copy of $U$ in $p^{-1}(U)$ is path connected, and $X_0$ is path connected, the disjoint sets of $p^{-1}(U)$ that intersect $X_0$ must lie entirely in $X_0$ (fill in details). Then, when we look at the restriction, $p\mid_{X_0}^{-1}(U)$ is a disjoint union of sets in $X$ each of which is mapped homeomorphically to $U$ by $p$, and therefore by $p\mid_{X_0}$, as required.
