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 ${\rho }_x=\sup \{r>0:B(x,r)\subseteq C_{\alpha }\text{ for some }\alpha \}\mathrm{.}$ Put $m_x=\min \{1,{\rho }_x\mathrm{/}4\}$ when ${\rho }_x<\mathrm{\infty }$, and $m_x=1$ otherwise. Then $0<m_x\le 1$ 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 $C_{\alpha }^{\mathrm{\prime }}$ 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 $\{C_{\alpha }^{\mathrm{\prime }}\}$ is an open refining cover.

Fix $x$. If $C_{\alpha }^{\mathrm{\prime }}$ meets $B(x,m_x\mathrm{/}8)$, choose a ball $B(y,m_y)\subseteq C_{\alpha }^{\mathrm{\prime }}$ meeting it. We claim $x\in C_{\alpha }$. Otherwise $x\mathrm{\notin }B(y,2m_y)$, while intersection gives $d(x,y)<m_y+m_x\mathrm{/}8$; hence $m_y<m_x\mathrm{/}8\le 1\mathrm{/}8$. 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 $d(x,y)+5m_y<6m_y+m_x\mathrm{/}8<7m_x\mathrm{/}8$; 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 $B(x,m_x\mathrm{/}8)$ meets only the corresponding finitely many $C_{\alpha }^{\mathrm{\prime }}$; hence the new cover is locally finite.

Hence $\{C_{\alpha }^{\mathrm{\prime }}\}$ 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.