Easier proof of "countable hypocompactness" I am interested in the following result, which appears as an old qual problem:

Let $X$ be a metric space and $\{U_i\}$ a countable open cover. Prove that there exists a countable open refinement $\{V_j\}$ such that for each $j$, we have $V_j\cap V_k\neq\emptyset$ for at most finitely many $k$.

In the literature, such refinements are called star-finite, and spaces with this property are called "countably hypocompact," although this language does not appear to be standard. I have proven the result more generally for paracompact Hausdorff spaces; but the proof (essentially due to Morita) is rather ungainly and seems too involved to be expected on a qualifying exam. It requires constructing a countable closed refinement of the original cover and performing a slightly intricate construction using Urysohn functions.
My question is this: is there a more direct proof in the case that $X$ is a metric space? I am tempted to exploit the existence of partitions of unity subordinate to the original cover, but alas star-finiteness is a stronger property than local finiteness (or point-finiteness, for that matter).
 A: $\newcommand{cl}{\operatorname{cl}}$There’s a fairly short but somewhat tricky proof for countably paracompact normal spaces. Let $X$ be such a space, and let $\mathscr{U}=\{U_k:k\in\Bbb N\}$ be a countable open cover of $X$. $X$ is countably paracompact, so $\mathscr{U}$ has a locally finite open refinement $\mathscr{R}$. For each $R\in\mathscr{R}$ choose $n(R)\in\Bbb N$ such that $R\subseteq U_{n(R)}$, and for $k\in\Bbb N$ let $V_k=\bigcup\{R\in\mathscr{R}:n(R)=k\}$; then $\mathscr{V}=\{V_k:k\in\Bbb N\}$ is a locally finite open refinement of $\mathscr{U}$ such that $V_k\subseteq U_k$ for each $k\in\Bbb N$.
Let $F_0=V_0\setminus\bigcup_{k>0}V_k$; $F_0$ is a closed subset of $V_0$, and $X$ is normal, so there is an open $W_0$ such that $F_0\subseteq W_0\subseteq\cl W_0\subseteq V_0$. Note that $\{W_0\}\cup\{V_k:k>0\}$ covers $X$. Suppose that $k>0$, and we have open sets $W_i$ for $i<k$ such that $\{W_i:i<k\}\cup\{V_i:i\ge k\}$ covers $X$, and $\cl W_i\subseteq V_i$ for each $i<k$. Let
$$F_k=V_k\setminus\left(\bigcup_{i<k}W_i\cup\bigcup_{i>k}V_i\right)\;;$$
$F_k$ is a closed subset of $V_k$, so there is an open $W_k$ such that $F_k\subseteq W_k\subseteq\cl W_k\subseteq V_k$. Clearly $\{W_i:i\le k\}\cup\{V_i:i>k\}$ covers $X$ so the construction goes through to give us a family $\mathscr{W}=\{W_k:k\in\Bbb N\}$ of open sets such that $\cl W_k\subseteq V_k$ for each $k\in\Bbb N$. Clearly $\mathscr{W}$ is locally finite, and the constructions ensures that it covers $X$.

On a qual one might be able simply to appeal to the very standard result that every point finite open cover of a normal space is shrinkable and avoid the construction of the previous paragraph. However, the real work is yet to come.

For $k\in\Bbb N$ let $W_k^*=\bigcup_{i\le k}W_i$ and $V_k^*=\bigcup_{i\le k}V_i$, and note that $\cl W_k^*\subseteq V_k^*$ for each $k\in\Bbb N$. By normality there are open sets $G_{k,\ell}$ for $k,\ell\in\Bbb N$ such that
$$\cl W_k^*\subset G_{k,\ell}\subseteq\cl G_{k,\ell}\subseteq G_{k,\ell+1}\subseteq V_k^*$$
for all $k,\ell\in\Bbb N$. For $n\in\Bbb N$ let $H_n=\bigcup\{G_{k,\ell}:k+\ell=n\}$; clearly $\{H_n:n\in\Bbb N\}$ is an open cover of $X$. Moreover, the construction ensures that $\cl H_n\subseteq H_{n+1}\subseteq V_n^*$ for each $n\in\Bbb N$: if $k+\ell=n$, then $\cl G_{k,\ell}\subseteq G_{k,\ell+1}\subseteq H_{n+1}\cap V_k^*\subseteq V_n^*$.
Let $B_0=H_0$ and $B_1=H_1$, and for $n\ge 2$ let $B_n=H_n\setminus\cl H_{n-2}$. Clearly each $B_n$ is open, and $B_n\subseteq H_n\subseteq V_n^*$ for each $n\in\Bbb N$. For $n\ge 2$ we have $\cl H_{n-2}\subseteq H_{n-1}$ and hence $$B_n=H_n\setminus\cl H_{n-2}\supseteq H_n\setminus H_{n-1}\;.$$ Checking the case $n=1$ separately, we see that in fact
$$H_n\setminus H_{n-1}\subseteq B_n\subseteq V_n^*$$
for all $n\ge 1$, and of course $H_0=B_0\subseteq V_0=V_0^*$, so $\{B_n:n\in\Bbb N\}$ covers $X$.
Finally, for each $n\in\Bbb N$ and $k\le n$ let $B_n(k)=B_n\cap V_k$, and let $$\mathscr{B}=\{B_n(k):k,n\in\Bbb N\text{ and }k\le n\}\;.$$
$B_n\subseteq V_n^*$, so $B_n=\bigcup_{k\le n}B_n(k)$, and $\bigcup\mathscr{B}=\bigcup_{n\in\Bbb N}B_n=X$, so $\mathscr{B}$ is an open cover of $X$. Evidently $\mathscr{B}$ refines $\mathscr{V}$ and hence $\mathscr{U}$, so it only remains to show that $\mathscr{B}$ is star-finite.
Fix $B_n(k)\in\mathscr{B}$; $B_n(k)\subseteq B_n\subseteq H_n$. If $m\ge n+2$, then $B_m=H_m\setminus\cl H_{m-2}\subseteq H_m\setminus H_n$, and $B_n(k)\cap B_m=\varnothing$. Thus, if $B_n(k)\cap B_m(\ell)\ne\varnothing$, then $\ell\le m\le n+1$, and we’re done.
I don’t at the moment see any way to use metrizability to simplify this. (This argument, by the way, is due to Kiyoshi Iséki in the short paper ‘A Note on Countably Paracompact Spaces’, Proc. Japan Acad. $30$ ($1954$), no. $5$, $350$-$351$.)
