Covering sequence of open subsets in a locally compact space This is from tom Dieck's notes on general topology:
https://www.uni-math.gwdg.de/tammo/algtop.html

Let the locally compact space $X$ be a union of compact subsets $(K_n)_{n \in \mathbb{N}}$. Then there exists a sequence $(U_n)_{n \in \mathbb{N}}$ of open subsets with the properties:

*

*$\overline{U_n}$ is compact for each $n \in \mathbb{N}$,


*$\overline{U_n} \subseteq U_{n + 1}$ for each $n \in \mathbb{N}$,


*$X = \bigcup_{n \in \mathbb{N}} U_n$.

The proof goes as follows. First, tom Dieck proves that for each compact subset $K$ of $X$ there can be chosen a compact set $t(K)$ such that $K \subseteq \mathrm{int}(t(K))$ (for this local compactness of $X$ is used). Then he recursively defined a sequence $(U_n)_{n \in \mathbb{N}}$ by setting $U_0 = \mathrm{int}(t(K))$ and $U_{n + 1} = \mathrm{int}(t(\overline{U_n}\cup K_{n+1}))$. This construction is only valid if we can show inductively that $\overline{U_n}$ is compact (as a finite union of compact subspaces is compact). That is, that $\overline{U_0} = \overline{\mathrm{int}(t(K))}$ is compact and $\overline{\mathrm{int}(t(\overline{U_n}\cup K_{n+1}))}$ is compact if $\overline{U_n}$ is. I can prove neither. One thing that imediately pops in mind a closed subset of a compact space is compact. However, I do not see how we can use it here.
 A: I doubt that it is true for non-Hausdorff locally compact $X$, The problem is that we cannot be sure that compact subsets are closed in that case.
So let us prove a modified theorem:
There exists a sequence of compact $C_n \subset X$ such that $C_n \subset \mathrm{int}(C_{n+1})$ and $\bigcup_n \mathrm{int}(C_n) = X$.
If $X$ is Hausdorff, then we can take $U_n = \mathrm{int}(C_n)$. Since $C_n$ is closed, we get $\overline{U_n} \subset C_n$. Because $\overline{U_n}$ is a  closed subset of a compact set, it is itself compact. This proves tom Dieck's theorem for Hausdorff locally compact $X$.
Now let us prove the above theorem.
Let $K$ be compact. Each $x \in K$ has a compact neighborhood $K(x)$. Finitely many $V(x_i) =  \mathrm{int}(K(x_i))$ cover $K$, thus $t(K)  = \bigcup K(x_i)$ is compact and clearly $V = \bigcup V(x_i)$ is an open set such that $K \subset V \subset t(K)$. This shows that $K \subset \mathrm{int}(t(K))$.
Now define $C_1 = t(K_1)$ and $C_{n+1} = t(C_n \cup K_{n+1})$. Then $C_n \cup K_{n+1} \subset \mathrm{int}(C_{n+1})$. In particular $C_n \subset \mathrm{int}(C_{n+1})$ and $K_n \subset \mathrm{int}(C_{n+1})$, thus $X = \bigcup K_n \subset \bigcup \mathrm{int}(C_{n+1}) = \bigcup \mathrm{int}(C_n)$. For the last equation recall $\mathrm{int}(C_1) \subset C_1 \subset \mathrm{int}(C_2)$.
