[proofplan]
Starting from a countable cover by compact sets (guaranteed by $\sigma$-compactness), we inductively construct an exhaustion $\{K_j\}$ where each $K_j$ is contained in the interior of $K_{j+1}$. At each stage, we enlarge the current compact set by covering it with finitely many precompact open neighbourhoods (provided by local compactness and the Hausdorff property) and taking the closure. The "moreover" statement follows from the nesting: any compact set $L$ is covered by the interiors $\operatorname{int}(K_j)$, which form an open cover of $X$, and compactness of $L$ extracts a finite subcover.
[/proofplan]
[step:Cover each compact set by finitely many precompact open neighbourhoods]
Since $X$ is $\sigma$-compact, write $X = \bigcup_{i=1}^\infty L_i$ where each $L_i \subset X$ is compact. Since $X$ is locally compact Hausdorff, for each $x \in X$ there exists an open set $U_x \subset X$ containing $x$ such that $\overline{U_x}$ is compact. (This uses local compactness to find an open neighbourhood contained in a compact set, and the Hausdorff property to ensure the compact set is closed by [Compact Subspaces and Hausdorff Spaces](/theorems/307), so that we may take $\overline{U_x}$ to be contained in that compact set, hence compact.)
For each compact set $L_i$, the open cover $\{U_x\}_{x \in L_i}$ admits a finite subcover by compactness. This observation is used in the inductive step below.
[/step]
[step:Inductively construct the exhaustion sequence]
We construct $\{K_j\}_{j=1}^\infty$ by induction. Set $K_1 := L_1$.
**Inductive step.** Suppose $K_j$ has been constructed and is compact. Consider the compact set $K_j \cup L_{j+1}$, which is compact as the union of two compact sets. For each point $x \in K_j \cup L_{j+1}$, choose a precompact open neighbourhood $U_x$ (as in the previous step). By compactness of $K_j \cup L_{j+1}$, finitely many of these cover it:
\begin{align*}
K_j \cup L_{j+1} \subset U_{x_1} \cup \cdots \cup U_{x_m}.
\end{align*}
Define
\begin{align*}
K_{j+1} := \overline{U_{x_1}} \cup \cdots \cup \overline{U_{x_m}}.
\end{align*}
Since each $\overline{U_{x_k}}$ is compact and a finite union of compact sets is compact, $K_{j+1}$ is compact. Moreover,
\begin{align*}
K_j \cup L_{j+1} \subset U_{x_1} \cup \cdots \cup U_{x_m} \subset \operatorname{int}\left(\overline{U_{x_1}} \cup \cdots \cup \overline{U_{x_m}}\right) = \operatorname{int}(K_{j+1}),
\end{align*}
where the last inclusion holds because $U_{x_1} \cup \cdots \cup U_{x_m}$ is open and contained in $K_{j+1}$, so it is contained in $\operatorname{int}(K_{j+1})$. In particular, $K_j \subset \operatorname{int}(K_{j+1})$.
[guided]
The inductive construction must achieve two goals simultaneously: (i) each $K_j$ sits inside the interior of $K_{j+1}$, and (ii) the union $\bigcup K_j$ exhausts $X$.
Goal (ii) is handled by incorporating $L_{j+1}$ at each step, ensuring that $L_{j+1} \subset K_{j+1}$ and hence $\bigcup K_j \supset \bigcup L_i = X$.
Goal (i) requires that $K_j$ is not merely contained in $K_{j+1}$ but in its interior. This is why we cover $K_j \cup L_{j+1}$ by finitely many open precompact sets $U_{x_1}, \ldots, U_{x_m}$ and define $K_{j+1}$ as the union of their closures. The open set $U_{x_1} \cup \cdots \cup U_{x_m}$ contains $K_j$ and is contained in $K_{j+1} = \overline{U_{x_1}} \cup \cdots \cup \overline{U_{x_m}}$. Since this open set lies inside $K_{j+1}$, it lies inside $\operatorname{int}(K_{j+1})$, giving $K_j \subset \operatorname{int}(K_{j+1})$.
The union of finitely many compact sets is compact (a finite union of finite subcovers is a finite subcover), so $K_{j+1}$ is compact. This completes the inductive step.
[/guided]
[/step]
[step:Verify the covering and absorption properties]
**Covering.** For each $i \in \mathbb{N}$, we have $L_i \subset K_i$ by construction (at step $i$, we ensured $L_i \subset K_i$). Therefore
\begin{align*}
X = \bigcup_{i=1}^\infty L_i \subset \bigcup_{j=1}^\infty K_j \subset X,
\end{align*}
giving $X = \bigcup_{j=1}^\infty K_j$.
**Nesting.** By the inductive construction, $K_j \subset \operatorname{int}(K_{j+1})$ for all $j \in \mathbb{N}$.
**Absorption of compact subsets.** Let $L \subset X$ be compact. The collection $\{\operatorname{int}(K_j)\}_{j=1}^\infty$ is an open cover of $X$: for any $x \in X$, there exists $j$ with $x \in K_j \subset \operatorname{int}(K_{j+1})$, so $x \in \operatorname{int}(K_{j+1})$. In particular, $\{\operatorname{int}(K_j)\}_{j=1}^\infty$ covers $L$. By compactness of $L$, finitely many suffice: $L \subset \operatorname{int}(K_{j_1}) \cup \cdots \cup \operatorname{int}(K_{j_N})$. Since the sequence $\{\operatorname{int}(K_j)\}$ is nested (because $K_j \subset \operatorname{int}(K_{j+1})$ implies $\operatorname{int}(K_j) \subset \operatorname{int}(K_{j+1})$), this reduces to $L \subset \operatorname{int}(K_J)$ where $J = \max(j_1, \ldots, j_N)$. In particular, $L \subset K_J$.
[/step]
