Let $(E,\tau)$ be a locally compact second-countable Hausdorff space. I want to show that there is a $(K_n)_{n\in\mathbb N_0}\subseteq E$ such that $K_n$ is compact and $K_{n-1}\subseteq\overset{\circ}{K_n}$ for all $n\in\mathbb N$ and $\bigcup_{n\in\mathbb N_0}K_n=E$.
The question is basically answered on mathoverflow, but there is a part of the proof that I don't understand.
Since $E$ is second-countable, $\tau$ has a countable basis $\mathcal B$. It's easy to see that $$\mathcal B_c:=\left\{U\in\mathcal B:\overline U\text{ is compact}\right\}$$ is again a basis of $\tau$.
Now, the construction described in the answer is as follows:
Let $U_0\in\mathcal B_c$ and $K_0:=\overline{U_0}$. Given $(U_0,K_0),\ldots,(U_{n-1},K_{n-1})$, let $U_n\in\mathcal B_c\setminus\left\{U_0,\ldots,U_{n-1}\right\}$, $\mathcal C$ be a finite subcover of $K_{n-1}$ and $K_n:=\overline{U_n}\cup\overline{C}$.
The problematic part is the cover $\mathcal C$ of $K_{n-1}$ (why subcover?). Can $\mathcal C$ really be arbitrary? What if we choose $\mathcal C=E$?
Best Answer
Clearly, it cannot be arbitrary. This is my interpretation:
As $\mathcal B_c$ is a basis of $E$, it is a covering of $K_{n-1}$. As $K_{n-1}$ is compact, there is a finite subcover $\mathcal C=\{B_1, B_2,..., B_n\}$ of $K_{n-1}$. Then $C=\bigcup B_i$ and $\overline C= \overline {B_1}\cup \overline {B_2}\cup...\cup \overline {B_n}$ is compact.