$\mathbf {The \ Problem \ is}:$ Show that any open covering of a locally compact metrizable space $X$ can be refined to a canonical covering . We say that a closed set $F$ of a topological space $X$ is called canonical if $F=\overline{\operatorname{Int}(F)}$ . A locally finite covering $\alpha=\{F_{\lambda}\}$ of a locally compact space $X$ is called a canonical covering if each $F_{\lambda}$ is a canonical closed set and $Int(F_{\lambda})\cap Int(F_{\delta})=\emptyset$ for all $\lambda\neq \delta.$
$\mathbf {My \ approach}:$ The details of canonical coverings as well as the solution of my problem is given in a paper "Homology Theory and the Exactness Axiom" by EG Skylarenko . I am attaching a screenshot as well as the link to the paper . At first , he takes a locally finite refinement
$\{U_{\lambda}\}$ of the open cover and that's possible because $X$ is metrizable . Then he orders the elements arbitrarily as $U_1,U_2,\cdots,U_{\lambda},\cdots$ but I can't find why that set can be ordered in such a way and after that he somewhat proceeds inductively . Does that mean the refinement has countably many elements ? The paper is here with page number $112.$
I am getting confused by this statement and need some leads . Thanks in advance for any help .
Skylarenko deals with an arbitrary locally compact metrizable space $X$; he definitely does not make a separability assumption.
For example, each uncountable discrete space is locally compact metrizable, and the open cover consisting of all singletons does not have a countable refinement.
Therefore it is clear that he must use transfinite induction.
What he needs is a well-ordering of the open covering $\alpha$. Then in particular $\alpha$ has a smallest element $U_1$, $\alpha \setminus \{U_1\}$ has a smallest element $U_2$ etc.
Let $R_\lambda : I(\lambda) \to \mathfrak C(X)$ be a function from the initial segment $I(\lambda) = \{ \mu \in \alpha \mid \mu < \lambda\}$ of $\alpha$ to the set $\mathfrak C(X)$ of canonical sets in $X$ such that the $F_\mu = R_\lambda(U_\mu)$ satisfy the appropriate conditions needed for a canonical covering. Skylarenko then extends $R_\lambda$ to $I(\lambda) \cup \{\lambda\}$ by $$F_\lambda = [U_\lambda \setminus \bigcup_{\mu < \lambda} F_\mu]$$ Observe that this already covers the case $\lambda = 1$ since $\bigcup_{\mu < \lambda} F_\mu = \emptyset$.
The principle of transfinite construction gives then a function $$R : \alpha \to \mathfrak C(X)$$ which is nothing else than the desired canoical covering.