Long line

Remarks

For each countable ordinal $\beta \ge 1$, the initial segment of $L$ below $(\beta, 0)$ is a countable stack of half-open intervals, and it is homeomorphic to $[0, 1)$. So every point of $L$ other than $(0, 0)$ has a neighbourhood homeomorphic to an open interval of $\R$. A sequence $(\alpha_n, t_n)$ in $L$ has a countable supremum $\gamma$ of the $\alpha_n$, because a countable union of countable sets is countable. So the sequence lies in the set of points up to $(\gamma + 1, 0)$, which is homeomorphic to $[0, 1]$, and it has a convergent subsequence. The open sets $\set{p \in L : p < (\alpha, 0)}$ for $\alpha < \omega_1$ cover $L$, but no finitely many of them do, because $\omega_1$ has no largest element. A sequentially compact metrizable space is compact, so $L$ is not metrizable. For the same reason the long line is a connected Hausdorff space, locally homeomorphic to $\R$, that is not second-countable and not metrizable. So being Hausdorff and locally Euclidean does not make a space a topological manifold in the usual sense, which also asks for second countability.