"Every open cover admits an open locally finite refinement" - can this refinement always be realized in terms of basis sets? Let $X$ be a paracompact topological space, let $C$ be an open cover, and let $\mathscr B$ be a basis for the topology. Does there always exist a locally finite refinement that consists of basis sets?
The reason this is trickier than it sounds is that, while every set in the refinement is a union of basis sets, this is definitely not guaranteed to be a locally-finite sort of union, and taking a locally finite refinement of this lands you outside the world of basis sets...
Now, let me add that I don't quite remember the context that led me to consider this question, but I'm still interested in finding out the answer.
 A: Interesting. Going back to my reading of Lee's Introduction to Topological Manifolds, I found the following theorem which I now realize is what prompted me to go here and ask this question which had been lingering in the back of my brain for a while. Note that this is not what made me originally consider the question.

Theorem 1.15 (Manifolds are Paracompact).
Every topological manifold is paracompact. In fact, given a topological manifold $M$, an open cover $\mathcal X$ of $M$, and any basis $\mathscr B$ for the topology of $M$, there exists a countable, locally finite open refinement of $\mathcal X$ consisting of elements of $\mathscr B$.

Here is the proof (as given by Lee):

Given $M$, $\mathcal X$, and $\mathscr B$ as in the hypothesis of the theorem, let $\left(K_j\right)_{j=1}^\infty$ be an exhaustion of $M$ by compact sets (Proposition A.60). For each $j$, let $V_j = K_{j+1}/\operatorname{Int}(K_j)$ and $W_j = \operatorname{Int}(K_{j+2}) / K_{j-1}$ (where we interpret $K_j$ as $\emptyset$ if $j < 1$). Then $V_j$ is a compact set contained in the open subset $W_j$. For each $x \in V_j$, there is some $X_x \in \mathcal X$ containing $x$, and because $\mathscr B$ is a basis, there exists $B_x \in \mathscr B$ such that $x \in B_x \subseteq X_x \cap W_j$. The collection of all such sets $B_x$ as $x$ ranges over $V_j$ is an open cover of $V_j$, and thus has a finite subcover. The union of all such finite subcovers as $j$ ranges over the positive integers is a countable open cover of $M$ that refines $\mathcal X$. Because the finite subcover of $V_j$ consists of sets contained in $W_j$, and $W_j \cap W_{j'} = \emptyset$  except when $j-2 \le j' \le j+2$, the resulting cover is locally finite $\blacksquare$

This seems to apply not only to manifolds, but also to second-countable locally-compact Hausdorff spaces (for which the Proposition A.60 holds, namely that they admit an exhaustion by compact sets). However, local compactness here seems key to making a locally finite refinement out of basis elements, so it seems like the statement may very well not hold in general. I have no idea how one might prove this one way or the other, though.
