Find a (simple?) counterexample to this statement about topological manifolds. Let us assume by a topological manifold $M$ of dimension $n$ I mean a Hausdorff topological space that is locally homeomorphic to $\mathbf{R}^n$, where $n$ is fixed.
I know that if $M$ is assumed separable and paracompact, then $M$ admits an exhaustion by compact sets.  
Is this still true if $M$ is only assumed separable, and not necessarily paracompact?
Thanks.

According to this MathOverflow answer, there are even non-normal separable complex manifolds; it gives a reference, and a comment below it notes that the version of the Prüfer surface mentioned in the cited reference is also separable and non-normal. And M. E. Rudin and P. Zenor, A perfectly normal non­-metrizable manifold, Houston.J. Math. $2$ ($1976$), $129­$-$134$, constructs a perfectly normal, hereditarily separable, non-metrizable manifold assuming $\mathsf{CH}$. If any of these admitted an exhaustion by compact sets, it would be second countable: each compact set is covered by finitely many charts, and each chart is second countable. But then it would be metrizable by the Uryson metrization theorem.
