Under choice, Ornstein's second construction turns a point-finite metric open cover into a locally finite open refinement

Well order the point-finite cover. For $x\in X$ let $\rho_x=\sup\{r>0:B(x,r)\subseteq C_\alpha \text{ for some }\alpha\}.$ Put $m_x=\min\{1,\rho_x/4\}$ when $\rho_x<\infty$, and $m_x=1$ otherwise. Then $0<m_x\le1$ and $B(x,2m_x)$ lies in some cover member: its radius is strictly below $\rho_x$. Assign $x$ to the first $C_\alpha$ containing $B(x,2m_x)$, and let $C'_\alpha$ be the union of all $B(x,m_x)$ assigned to $\alpha$.

The selected smaller balls cover $X$ and each lies in its assigned $C_\alpha$, so $\{C'_\alpha\}$ is an open refining cover.

Fix $x$. If $C'_\alpha$ meets $B(x,m_x/8)$, choose a ball $B(y,m_y)\subseteq C'_\alpha$ meeting it. We claim $x\in C_\alpha$. Otherwise $x\notin B(y,2m_y)$, while intersection gives $d(x,y)<m_y+m_x/8$; hence $m_y<m_x/8\le1/8$. Thus the truncation in step 1.1 is inactive at $y$ and $\rho_y=4m_y$. But $B(y,5m_y)\subseteq B(x,2m_x),$ because $d(x,y)+5m_y<6m_y+m_x/8<7m_x/8$; the right-hand ball lies in some cover member by step 1.1. This contradicts the definition of $\rho_y$. Thus $x\in C_\alpha$.

The input cover is point-finite, so $x$ belongs to only finitely many $C_\alpha$. Step 2.2 shows that $B(x,m_x/8)$ meets only the corresponding finitely many $C'_\alpha$; hence the new cover is locally finite.

Hence $\{C'_\alpha\}$ is a locally finite open refinement of the point-finite cover. For an arbitrary metric open cover, first apply [L1] and then this construction; refinement is transitive, so the result refines the original cover.