the equivalence between paracompactness and second countablity in a locally Euclidean and $T_2$ space suppose $M$ is a locally Euclidean Hausdorff space, show that $M$ is second countable if and only if it is paracompact and has countably many components.
 This is Problem 2-15 p.59 (or 1-5 p.30 in the new version) in: Introduction to smooth manifolds by John M. Lee.
In the hint he said if $M$ is paracompact, show that each component of $M$ has a locally finite cover by precompact open balls, and extract a subcover. But since we don't know about the metrizability of the space, every open cover has a subcover doesn't necessarily mean the space is second countable? 
Am I missing something here?
 A: First of all, a topological space that is locally homeomorphic to a Euclidean space is locally compact, cf. this, and locally connected as those properties are local and preserved by homeomorphisms (and they hold for Euclidian spaces). Then
$M$ second-countable $\Longrightarrow\ M$  paracompact with countably many connected components:
This is Lemma 1.9 p.9 Foundations of Differentiable Manifolds and Lie Groups, Frank W. Warner or Thm 4.77 p.110 Introduction to Topological Manifolds, John Lee:
The authors show in the first part that $M$ is countable at infinity/$\sigma$-compact, i.e. $M = \bigcup_{n\in \mathbb{N}} K_n$ with $K_n \subset \overset{\circ}{K}_{n+1}$ compact. 
Let now $ \mathcal{U}:=(U_i)_{i\in I}$ be any open cover of $M$:
$K_0$ is compact and covered by the family open subsets $(U_i\cap \overset{\circ}{K}_1)_{i\in I}$ so one can extract a finite subsequence. (Similarly for $K_1$)
Then by induction: for any $n \geq 2,\enspace \left(U_i\cap (\overset{\circ}{K}_{n+1}\backslash \overline{K}_{n-2})\right)_{i\in I}$ is an open cover of $\overline{K}_n\backslash K_{n-1}$, from which one extracts a finite cover.
By such a construction, one will cover every $K_n$ and hence all of $M$; the open subset of the cover are of the form $U_i\cap (\overset{\circ}{K}_{n+1}\backslash \overline{K}_{n-2})$ and thus included in $U_i$ (refinement of a cover) and any point is in one of the $\overset{\circ}{K}_n$ which is covered by finitely many $U_i$ (locally finite cover), i.e. $M$ is paracompact.
Edit: this last implication is not well justified here because a cover of $K_{n+1}$ or $K_{n-1}$ may also intersect with points of $K_n$ ... although the intuition is correct...
In this post, one uses the fact that a second countable space is separable, together with local connectedness to show that $M$ has countably many connected components.
$M$  paracompact with countably many connected components $\Longrightarrow\ M$ second-countable :
Since $M$ has countably many connected components $(C_k)_{k\in \mathbb{N}}$, it suffices to show that each of these components has a countable basis ($\mathbb{N}\times \mathbb{N}$ is still countable).
The idea is to use the fact that for any chart $(U,\varphi)$ (from definition of $M$ locally Euclidian), $ U$ is second-countable as it is homeomorphic to a subset of $\mathbb{R}^n$ which is second-countable.
One shows again that $M$ is countable at infinity / $\sigma$-compact but starting with the hypothesis that $M$ paracompact. It is similar as above, cf. Theorem 12.11 p.38 of Topology and Geometry, Glen E. Bredon.
Finally with $\sigma$-compactness, $M$ is covered by countably many open subsets, each of which are second-countable.

I found these notes which also answers precisely this question link
