a covering map is open? $E,B$ are topological spaces and let's say that $p:E\to B$ is a covering map.
Is $p$ open?
i tried to show it as follows:
let $U$ be an open set in $E$, and now for every $x\in U$, $p(x)\in B$. $p$ is a covering map so there is a open neighborhood of $p(x)$, call it $V$ that is evenly coverd by $p$. e.g there is a partition of $p^{-1}(U)$ of open disjoint subsets $\{V_\alpha\}$ that covers $E$. and $p|_{V_\alpha}$ is homeomorphism between $V_\alpha$ and $V$.
so for some $\alpha$, $x\in V_\alpha$. so $V_\alpha \bigcap U$ is an open neighborhood of $x$ and $V_\alpha \bigcap U\subset V_\alpha$ and it is open in $V_\alpha$
, and since $p|_{V_\alpha}$ is homeomorphism between $V_\alpha$ and $V$ we get that $p(V_\alpha \bigcap U)$ is an open neighborhood of $p(x)$ and it is a subset of $p(U)$.
then $p(U)$ is open.
is it right?

Your argument is quite correct, but start with let $y = p(x)$ be a point of $p[U]$ where $x \in U$, and then proceed (you start with an arbitrary element of $p[U]$ and show its an interior point, not an element of $U$, though here the difference is slight...).
This generalises to local homeomorphisms $f: X \rightarrow Y$, where every $x$ has an open neighbourhood $U_x$ such that $f|_{U_x}$ is a homeomorphism between $U_x$ and the open set $f[U_x]$. The argument is almost the same.
