Connected, locally compact, paracompact Hausdorff space is exhaustible by compacts I'm trying to understand the proof of the following:

A connected, locally compact, paracompact Hausdorff space $X$ has an exhaustion by compact sets,

That is, there exists a sequence $(K_n)_n$ of compact subsets of $X$ whose union is $X$ itself, such that $K_n$ is included in the interior of $K_{n+1}$, for all $n$.
The proof goes as follows. Choose a locally finite open cover $(U_i)_{i∈I}$ of $X$, such that the closure of $U_i$ is compact, for all $i$. Then every compact subset of $X$ intersects only finitely many of the $U_i$ (why?).
Then $K_1$ is chosen as the closure of any nonempty $U_i$. $K_2$ is chosen as the union of the closures of the $U_i$ that intersect $K_1$ and so on. The rest is easy.
 A: One starts by noting that all the set of all open $O$ with $\overline{O}$  compact is an open cover of $X$ by locally compactness and Hausdorffness.
The paracompactness of $X$ then gives us a locally finite refinement $(U_i)_{i \in I}$ of that cover. It's not the $U_i$ that need to be compact in this proof, but their closures and this follows as each $U_i$ is a subset of some $O$ with compact closure so the same holds for the $U_i$. The new thing is the local finiteness, which is used for the compact set fact:
Now if $K$ is compact, each $x \in K$ has a neighbourhood $W_x$ such that $\{i \in I: U_i \cap W_x\neq \emptyset \}$ is finite, as we have a locally finite refinement. 
Then $K$ being compact is covered by finitely many of these $W_x$, say $W_x, x \in F$ for some finite $F \subseteq K$. 
But then $$\{i \in I: K \cap U_i \neq \emptyset \} \subseteq \{i \in I: ( \bigcup_{x \in F} W_x ) \cap  U_i \neq \emptyset\} = \bigcup_{x \in F} \{i \in I: W_x \cap U_i\neq \emptyset\}$$ where the latter is a finite union of finite sets so finite.
So the $\{U_i: i \in I\}$ are as needed.
A: Suppose $K\subseteq X$ is compact.  For each $x\in K$, choose an open set $V_x$ around $x$ that intersects only finitely many $U_i$ (by local finiteness).  Since $K$ is compact, it is covered by only finitely many of these $V_x$.  Since each $V_x$ intersects only finitely many $U_i$, so does $K$.
