Let $p: E\rightarrow B$ a continuous surjective map and $U \subseteq B$ be open and not empty and who is being evenly covered by $p$. Show that all non-empty subsets of $U$ are being evenly covered by $p$.
What I've got so far with the hints:
Let $W \subseteq U$ with $W$ open and non-empty.
$$ W = W \cap U = W \cap p( \cup V_\alpha) $$
So that $$ p^{-1}(W) = p^{-1}(W \cap p(\cup V_\alpha)) = p^{-1}(W) \cap \cup V_\alpha = \cup (p^{-1}(W) \cap V_\alpha) $$
Since $p$ is continuous and $W$ is open in $B$ we know $p^{-1}(W)$ is open and $V_\alpha$ is open by definition. So $p^{-1}(W) \cap V_\alpha$ is open.
Next up we have to prove that:
$$ p: p^{-1}(W) \cap V_\alpha \rightarrow W $$ is a homeomorphism. By definition $p$ is continuous and surjective and notice that only $p(0) = 0$ so this map is also injective. If $Z \subseteq W \subseteq U$ with $Z$ open then we know
$$ p(p^{-1}(Z) \cap V_\alpha) = Z \cap V_\alpha $$ is open. So $p$ is also an open mapping and thus is $p$ a homeomorphism and thus is $W \subseteq U$ being evenly covered by $p$.
Since $W\subseteq U$, you have $p^{-1}(W)=p^{-1}(W)\cap\coprod_\alpha V_\alpha$ which is the union $\coprod_\alpha p^{-1}(W)\cap V_\alpha$. Can you show that each $p^{-1}(W)\cap V_\alpha$ is open and that $p:p^{-1}(W)\cap V_\alpha\to W$ is a homeomorphism?