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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?

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    Cover $M$ by a countable family of evenly covered open sets. (That's possible because $M$ is second countable.) For each of these sets, the preimage is a countable disjoint union of copies of the base set, and that makes it second countable. The union of countably many countable families is countable. – 2017-02-01
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    I am not sure if I understand how does second countability of $M$ impy that it can be covered by a countable family of evenly covered opet sets since second countability just refers to a countable basis. The rest I get – 2017-02-01
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    Take a countable basis of the topology, and take the subfamily of evenly covered sets in that basis. Clearly that's a countable family. It also covers $M$. – 2017-02-01
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    You mean take only those evenly covered sets that are contained in basis? But how does that cover $M$? – 2017-02-01

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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.

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    That was precise and useful. Thanks a lot! – 2017-02-01