If a locally finite open cover has locally finite closed refinement, does it have a locally finite open barycentric refinement? We can get an open barycentric refinement like this.
Let $(U_i)_{i \in I}%$ be an locally finite open cover of $X$, and $(F_j)_{j \in J}$ be a locally finite closed refinement of it. Pick for every $j \in J$ $i(j)$ such that $F_j \subset U_{i(j)}$. Let $V_x = \bigcap \{ U_{i(j)} \mid x \in F_j \} \setminus \bigcup \{ F_j \mid x \not\in F_j \}$. Since $x \in V_x$, $(V_x)_{x \in X}$ is an open cover of $X$ and if $x \in V_y$ and $x \in F_j$, $y \in F_j$ thus $V_y \subset U_{i(j)}$ implying that $(V_x)_{x \in X}$ is an open barycentric refinement of $(U_i)_{i \in I}$.
The book states that if $(U_i)_{i \in I}$ is locally finite, then $(V_x)_{x \in X}$ is also locally finite (maybe after removing duplicates). Let $x \in V$ such that $V$ meets $U_i$ only when $i \in I'$ for some finite set $I' \subset I$. For $V$ to meet $V_y$, $\{i(j) \mid y \in F_j \} \subset I'$ should hold. The problem is there can be infinitely many $j$s having same $i(j)$ so this is insufficient to establish the local finiteness of $(V_x)_{x \in X}$. How can I proceed?
This fact is proved in Theorem 1 in my note here (under $(2)$). I refer to the proof there as then the notation is consistent (it's a bit different than yours, though it's otherwise the same idea).
In your notation: let $x \in X$ and let the open neighbourhood $O_x$ of $x$ be a witness to the local finiteness of$(U_i)_{i \in I}$ at $x$, so that $I(O_x) = \{i \in I\mid O_x \cap U_i \neq \emptyset\}$ is finite. Now consider $O_x \cap V_x$: suppose that $(O_x \cap V_x) \cap V_p \neq \emptyset$ for some $p \in X$. Let $j \in J$, then if $p \in F_j$, then $V_p \subseteq U_{i(j)}$ by the definition of $V_p$ and so $i(j) \in I(O_x)$. It follows that the $i(j)$ and $j$ used in $V_p$ can only be that correspond to subsets of $I(O_x)$ of which there are only finitely many.
So $O_x \cap V_x$ witnesses the local finiteness of the refinement cover at $x$.