Let $E,X$ be topological spaces and $\pi:E \to X$ a covering map. Show that $\pi$ is a local homeomorphism. 
Let $E,X$ be topological spaces and $\pi:E \to X$ a covering map. Show that $\pi$ is a local homeomorphism.

Since $\pi$ is a covering map we have that for every $p \in X$ there exists a neighborhood $U$ such that $\pi^{-1}(U)$ is a union of disjoint open sets in $E$, each of which is mapped homeomoprhically to $X$.
The definition of local homeomorphism states that in order for $\pi$ to be such a map we need that for every $x \in E$ there exists open set $O$ such that $x \in O$ and $\pi(O)$ is open in $X$ and the restriction $\pi\mid_U : U \to \pi(U)$ is a homeomorphism.
I don't quite now how to approach this. If I pick $x \in E$, then $x=\pi^{-1}(y)$ for some $y \in X$, but for every $y \in X$, there exists nbhd $U$ containing $y$ such that $\pi^{-1}(U)$ is the union of disjoint open sets in $E$ that are mapped homeomoprhically to $X$.
How can I use this info to show the requirements for this to be locally homeomorphic?

If $p \in X$ let $y=p(x)$ and let $U$ be an evenly covered neighbourhood of $y$ which means that $p^{-1}[U] = \bigcup_{i \in I} O_i$ (disjoint union) so that for all $i$, $p \restriction_{O_i}: O_i \to U$ is a homeomorphism.
One of the $O_i$ contains $x$ so you're done.
