A covering space of a manifold is a manifold I have been told that for a given covering map $ξ: M' → M$ and under the assumtion that $M$ is a manifold, his covering space $M'$ is again manifold if we assume at most countable fibre. 
I have proven everything except for the second countability part for which I suppose most_countable_fibre assumption should be used, but I got stuck in proving how. 
Can anyone explain me why?
 A: Let $\mathscr{B}$ be a countable basis of the topology of $M$, and
$$\mathscr{E} = \{ U \in \mathscr{B} : U \text{ is evenly covered}\}.$$
$\mathscr{E}$ is countable as a subset of a countable set, and $\mathscr{E}$ covers $M$: if $x\in M$, then by definition of a covering there is an open $V\subset M$ containing $x$ such that $V$ is evenly covered. Since $\mathscr{B}$ is a basis of the topology, there is a $U\in \mathscr{B}$ with $x \in U \subset V$. Subsets of evenly covered sets are evenly covered, so $U \in \mathscr{E}$. Since $x\in M$ was arbitrary, we have shown that $\mathscr{E}$ covers $M$.
For $U \in \mathscr{E}$, we have
$$\xi^{-1}(U) \cong U \times D_U,$$
where $D_U$ is a countable (finite or countably infinite) discrete space. Since $U$ and $D_U$ are second countable, so is $U \times D_U$, hence $\xi^{-1}(U)$. Let $\mathscr{B}_U$ be a countable basis of the topology of $\xi^{-1}(U)$. Since the $\xi^{-1}(U)$ are open,
$$\mathscr{B}_{\mathscr{E}} = \bigcup_{U \in \mathscr{E}} \mathscr{B}_U$$
is a basis of the topology of $M'$. As a countable union of countable sets, it is countable. Hence $M'$ is second countable.
