Second-countable locally compact Hausdorff spaces admit compact exhaustions

Fix an at most countable basis $\mathcal{B}$ for $S$ and let $\mathcal{U}=\{U\in \mathcal{B}:\bar{U}\text{ is compact}\}$. For any $x\in S$, [F2] with $O=S$ gives an open $V$ containing $x$ whose closure is compact. The basis property gives $U\in \mathcal{B}$ with $x\in U\subseteq V$; then $\bar{U}\subseteq \bar{V}$, and $\bar{U}$ is closed in the compact space $\bar{V}$, hence compact by [F3]. Thus $\mathcal{U}$ is an at most countable open cover of $S$.

If $S=\mathrm{\varnothing }$, take $K_n=\mathrm{\varnothing }$. Otherwise enumerate the subfamily $\mathcal{U}$ by filtering the fixed countable basis in its given enumeration; if the subfamily is finite, repeat its last member. Put $K_n={\bigcup }_{j=1}^n{\bar{U}}_j$. An open cover of $K_n$ restricts to an open cover of each of the finitely many compact sets ${\bar{U}}_j$; taking the union of their finite subcovers gives a finite subcover of $K_n$. Thus $K_n$ is compact. The sequence is increasing, and its union is $S$ because $\mathcal{U}$ covers $S$.

Let $C\subseteq S$ be compact. The open cover $\mathcal{U}$ of $C$ has a finite subcover by the ambient-cover criterion in [F4]. Let $n$ be at least every index in that finite subcover. Then $C\subseteq {\bigcup }_{j=1}^n{U}_j\subseteq K_n$. Thus every compact $C$ is contained in some $K_n$.

Steps 2.1 and 3.1 give the required increasing compact exhaustion and its cofinal containment property. The construction uses only a fixed countable basis and finite subcovers, so it invokes no choice principle.