Proof of some properties of a locally compact separable metric space I want to find a proof of the following theorem:

Let $X$ be a topological space. The following two statements are
equivalent:

*

*$X$ is a locally compact, second-countable, Hasudorff space

*$X$ is a locally compact, separable, metric space

Moreover, if the above equivalent conditions are satisfied, then we
have:

*

*$X$ is completely metrizable.

*Every open set has an exhuastion by compact sets. More precisely, if
$U \subseteq X$ is an open set, then there exists a nested sequence
of compact subsets $ \{K_n\}_{n=1}^\infty $ in $U$ such that

$$ \forall n, K_n \subseteq \mathrm{int}(K_{n+1}) \quad \text{and}
 \quad U = \bigcup_{n=1}^\infty K_n $$

I know that the equivalence of the two conditions follows from the Urysohn metrization theorem and the fact that separable metric space is second-countable. But I have no idea how to prove the remaining parts of the theorem. Could anyone give a rigorous proof?
 A: As to the equivalence: that is indeed a simple consequence of Urysohn and some basic facts:

*

*suppose that $X$ is locally compact, second countable and Hausdorff.

*Then $X$ is regular $T_1$ (loc. cpt plus Hausdorff implies that)

*So Urysohn implies $X$ is locally compact separable metrisable. (of course second counatble always implies separable).

To go back is trivial as a separable metric space is second countable and all metric spaces are Hausdorff.
Now hints to the final properties:
Such $X$ can be given a complete metric: first note that the one-point compactification $\alpha X =X \cup \{\infty\}$ is compact Hausdorff and also second countable (hint: construct a countable local base at $\infty$ by considering a countable base on $X$ by open sets with compact closure).
So $\alpha X$ is complete metric (Urysohn again gives a metric and on a compact space any compatible metric is complete).
By local compactness $X$ is open in $\alpha X$, and it's well-known that a $G_\delta$ in a a compact metric space is completely metrisable.
For the exhaustion in 2. Let $\mathcal{B}=\{B_n\mid n \in \Bbb N\}$ be a countable base of $X$ so that all $\overline{B_n}$ are compact. Now define the $K_n$ by a simple recursion starting with $K_1 = \overline{B_0}$...
A: We shall use the following property:

If a space $X$ is locally compact, Hausdorff and second-countable,
then it has a countable base, each member of which has a compact
closure.

Let $\mathscr{B}$ be a countable base of $X$. For each $x\in X$, pick a compact neighbourhood $K$ of $x$, and let $\mathscr{B}_x$ be the collection of all members of $\mathscr{B}$ contained in $K$. The union $\mathscr{V}$ of all $\mathscr{B}_x$ is a compact base (in the sense that the closure of each base set is compact).
With the property established, now let $(X,d)$ be a locally compact separable metric space. We may assume that $0\leq d\leq 1$.
We prove 2 first. Let $\mathscr{V}$ be as above, and $\mathscr{V}'=\{V_1,V_2,\dots\}$ be the collection of all members of $\mathscr{V}$ whose closures are contained in $U$. Since $\mathscr{V}$ is a base, $U=\bigcup_{n=1}^\infty V_n$.
Given positive integers $m,n$, put
\begin{align}
G_{m,n}=\{x\in X\colon d(x,X-V_n)\geq1/m\}.
\end{align}
Then each $G_{m,n}$ is compact, and $G_{m,n}\subseteq G^o_{m+1,n}$ ($^o$ stantds for interior). Moreover, since $V_n$ is open, $V_n=\bigcup_{m=1}^\infty G_{m,n}$. The sets
\begin{align}
K_n=G_{n,1}\cup\dots\cup G_{n,n}
\end{align}
are as required.
To prove 1, Let $K_1,K_2,\dots$ be as in part 2, where $U$ is replaced by $X$. For each $n$, let $U_n$ be the interior of $K_{n+1}$, and define $f_n\colon X\to[0,1]$ by
\begin{align}
f_n(x)=\frac{d(x,X-U_n)}{d(x,K_n)+d(x,X-U_n)}.
\end{align}
Then $f_n$ is $1$ on exactly $K_n$ and $0$ on exactly $X-U_n$. Put \begin{align}
f=\sum_{n=1}^\infty 2^{-n}f_n.
\end{align}
The convergence is uniform, so $f$ is continuous. Moreover, $f(x)<2^{-n}$ if and only if $x\notin K_{n+1}$.
Define a metric $\rho$ on $X$ by
\begin{align}
\rho(x,y)=d(x,y)+\left|\frac{1}{f(x)}-\frac{1}{f(y)}\right|.
\end{align}
Fix $x\in X$. Given $r>0$, there exists $\delta\in(0,r/2)$ such that $d(x,y)<\delta$ implies
\begin{align}
\left|\frac{1}{f(x)}-\frac{1}{f(y)}\right|<r/2.
\end{align}
Then $\rho(x,y)<r$ whenever $d(x,y)<\delta$. On the other hand, $d(x,y)\leq\rho(x,y)$ for all $y\in X$. Hence $d$ and $\rho$ induce the same topology on $X$.
Let $(x_n)_{n\geq 1}$ be a $\rho$-Cauchy sequence. Pick $\epsilon>0$, there exists $N$ such that
\begin{align}
\left|\frac{1}{f(x_N)}-\frac{1}{f(x_n)}\right|<\epsilon.
\end{align}
for all $n\geq N$. Then $1/f(x_n)$ cannot be arbitrarily large, so the sequence $(x_n)$ is in $K_i$ for some positive integer $i$. Since $K_i$ is compact, the sequence is convergent. Hence $(X,\rho)$ is complete.
