Question: This question arose from considering for a connected smooth Hausdorff manifold the (possible) equivalence of the following properties:

(1) paracompact,  

(2) metrizable,  

(3) second countable,  

(4) countable at infinity,  

(5) $\sigma$−compact,  

(6) Lindelöf,  

(7) separable.

I know proofs for the equivalence of the first six, and that they imply (7), but it is problematic, whether this implies the others. By *countable at infinity* I mean existence of a sequence of compact sets $K\_i$ whose union covers the space and which satisfies $K\_i\subseteq{\rm Int\ }K\_{i+1}$ . Of course, *locally euclidean* means that each point has a neighbourhood homeomorphic to $\mathbb R^k$ with the standard topology and $k\in\mathbb N$ .

Answer: 
I answer my our question: *Separability of a connected locally euclidean Hausdorff topological space **does not** imply second countability*, or any of the equivalent conditions (1), ... (6) given in the question. A counterexample is given in Example 5 on page 15 in 
[David Gauld's preprint](http://arxiv.org/PS_cache/arxiv/pdf/0910/0910.0885v1.pdf). There is constructed a separable Hausdorff topological space, which is not second countable. The space can be equipped with a compatible **analytic** atlas modelled on $\mathbb R^2$ . One such is $\lbrace{\rm id\ }S\rbrace\cup\lbrace\phi\_{\eta,\zeta}:\eta,\zeta\in\mathbb R\rbrace$ , where $\phi\_{\eta,\zeta}$ is given by $(0,\eta,z)\mapsto(0,z-\zeta)=(0,v)$ when $|v|<1$ , and $(x,y)\mapsto(x,|x|^{-1}(y-\eta)-\zeta)=(u,v)$ when $0<|x|<1$ and $|v|<1$ . So Gauld's space **does not** satisfy any of the conditions (1), ... (6).
Indeed, check [the paper by Gauld](http://arxiv.org/pdf/0910.0885v1). Your (4) implies his condition hemicompact.
His example at p15, that you saw refutes the just separable condition (7). Note that (1)-(6) imply imply metrisability for just continuous manifolds, so it still might be that the situation vis à vis separability is different for **smooth** manifolds instead of continuous ones, though I suspect not.