I was reading "Munkres, Topology 2ed, p338-339". It is about theorem 53.2. The last sentence of the proof, '$V_\alpha\cap E_0$ is mapped homeomorphically onto $U\cap B_0$ by $p$' is so implicit that I can't touch the detailed stuff.
So I made my own proof below, which uses only elementary facts. In my proof, there are two questionmarks in (4) and (5), at which I used an equation that is not generally accepted : $f(A)\cap f(B)=f(A\cap B)$. Sorry for a complicated proof, but please let me know that my proof is valid or how I can modify it.
[Definition, evenly covered] Suppose $p:E\to B$ is surjective and continuous. An open set $U$ in $B$ is evenly covered by $p$ if $p^{-1}(U)$ is a disjoint union of $V_\alpha$ and $V_\alpha\cong U$(homeomorphic by $p$) for each $\alpha$.
[Definition, covering map] Suppose $p:E\to B$ is surjective and continuous. $p$ is said to be a covering map if, for each $b\in B$, there exists a neighborhood $U$ that is evenly covered by $p$.
[Theorem] Let $p:E\to B$ be a covering map. If $E_0=p^{-1}(B_0)$, (said differently, $E_0$ is saturated) then the restriction $p_0=p|_{E_0}:E_0\to B_0$ is a covering map.
[My proof] Suppose $b_0\in B_0$. Since $p:E\to B$ is a covering map and $b_0\in B$, there exists a neighborhood $U$ of $b_0$ that is evenly covered. So there exists a collection of disjoint open sets $\{V_\alpha\}$ such that $$\bigcup_\alpha V_\alpha=p^{-1}(U)\tag{1}$$ and $$V_\alpha\cong U(\text{homeomorphic by $p$}).\tag{2}$$
Now consider $B_0\cap U$ and $\{E_0\cap V_\alpha\}$. $B_0\cap U$ is an open set in $B_0$ which contains $b_0$, so it is a neighborhood of $b_0$. Note also that for each $\alpha$, $E_0\cap V_\alpha$ is disjoint one another. It follows from (1) and $E_0=p^{-1}(B_0)$ that \begin{align*} E_0\cap\bigcup_\alpha V_\alpha&=p^{-1}(B_0)\cap p^{-1}(U)\\ \bigcup_\alpha(E_0\cap V_\alpha)&=p^{-1}(B_0\cap U). \end{align*}
In order for $B_0\cap U$ be evenly covered by $p_0$, it suffices to prove that $$E_0\cap V_\alpha\cong B_0\cap U(\text{homeomorphic by }p_0).\tag{3}$$ The fact that $p_0$ is continuous and injective is trivial since $p_0$ is a restriction of a continuous and injective map $p$. Surjectivity of $p_0$ is immediate from $$p_0(E_0\cap V_\alpha)=p(E_0\cap V_\alpha)\stackrel?{=}p(E_0)\cap p(V_\alpha)\stackrel{(2)}=B_0\cap U.\tag{4}$$ It remains to prove that $p_0$ has a continuous inverse, that is, $p_0$ is an open map.
Consider an open set $E_0\cap V_\alpha\cap W$ in $E_0\cap V_\alpha$ where $W$ is open in $E$. Since $V_\alpha\cap W$ is open in $V_\alpha$, (2) implies that $p(V_\alpha\cap W)$ is open in $U$. So $p(V_\alpha\cap W)=T\cap V$ where $T$ is open in $B$. Thus $$p_0(E_0\cap V_\alpha\cap W)=p(E_0\cap V_\alpha\cap W)\stackrel?=p(E_0)\cap p(V_\alpha\cap W) =B_0\cap T\cap U\tag{5}$$ is an open set in $B_0\cap U$. Therefore, $p_0$ is indeed an open map, $B_0\cap U$ is evenly covered by $p_0$, and $p_0$ is an open mapping.
Your proof is correct, but why do you want to ignore the fact that the $p_\alpha : V_\alpha \stackrel{p}{\rightarrow} U$ are homeomorphisms? You correctly prove that $p_0^{-1}(B_0 \cap U) = \bigcup_\alpha E_0 \cap V_\alpha$, where the $V'_\alpha = E_0 \cap V_\alpha$ are pairwise disjoint open subsets of $E_0$. But $p_0 \mid_{V'_\alpha} = p \mid_{V'_\alpha} = p_\alpha \mid_{V'_\alpha}$, thus $p_0 \mid_{V'_\alpha}$ is the restriction of the homeomorphism $p_\alpha$ and thus maps $V'_\alpha$ homeomorphically onto $p_\alpha(E_0 \cap V_\alpha) = B_0 \cap U$.