Does locally compact separable Hausdorff imply $\sigma$-compact? Is it true that every locally compact separable Hausdorff space is $\sigma$-compact?

The following is counterexample 65 in Counterexamples in Topology, the rational sequence topology. Let $X=\mathbb{R}$ and for each $r\in\mathbb{R}$, pick a sequence $(q_n^r)$ of rationals converging to $r$. Let the basis of the topology be all sets of the form $\{q_n^r,q_{n+1}^r,\ldots\}\cup\{r\}$ for some $r$ and $n$ and all sets of the form $\{q\}$ with $q$ rational. The space is separable, since every basic open set contains a rational number. The space is Hausdorff, since sequences converging to different points are eventually in disjoint neighborhoods. The space is locally compact, since $\{q_1^r,q_2^r,\ldots\}\cup\{r\}$ is a compact neighborhood of $r$. But a compact set can contain only finitely many irrational numbers, and since the set of irrational numbers is uncountable, the space is not $\sigma$-compact.
