Does locally compact separable Hausdorff imply $\sigma$-compact?

It’s not true in general. Let $\mathscr{A}\,$ be an uncountable family of pairwise almost disjoint infinite subsets of $\omega$ (i.e., distinct members of $\mathscr{A}\,$ have finite intersections). Let $X=\omega\cup\mathscr{A}\,$. Points of $\omega$ are isolated, and for each $A\in\mathscr{A}\,$, $$\Big\{\{A\}\cup (A\setminus F):F\subseteq \omega\text{ is finite}\Big\}$$ is a local base at $A$. ($X$ is an example of a Mrówka $\Psi$-space.) It’s easy to see that $X$ is Tikhonov, zero-dimensional, separable, and locally compact, but $$\Big\{\{A\}\cup\omega:A\in\mathscr{A}\,\Big\}$$ is an open cover of $X$ with no countable subcover, so $X$ is not Lindelöf. However, a locally compact space is Lindelöf iff it’s $\sigma$-compact, so $X$ cannot be $\sigma$-compact. (Of course it’s easy enough to see directly that $X$ is not $\sigma$-compact: if $X=\bigcup_n Y_n$, some $Y_n\cap\mathscr{A}\,$ must be infinite, and then $Y_n$ cannot be 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.