Why is paracompactness needed to prove a regular space is normal?

I am trying to understand why paracompactness is needed to prove a regular space is normal. It was used in the prove above to find a locally finite open refinment $\{w_\lambda\}$ from an open cover. But I fail to understand why we need that. I mean why don't we just set $V=⋃_{x\in B} V_x , U=X| ⋃_{x\in B} C(V_x)$. Why do we need the locally finite open refinment $\{w_\lambda\}$?

Solution 1:

Local finiteness is used in the step where they say $U=X\setminus\bigcup_\gamma\overline{W_\gamma}$ is open since $\bigcup_\gamma \overline{W_\gamma}=\overline{\bigcup W_\gamma}$. The latter equality is not true for an arbitrary collection of sets $(W_\gamma)$, but it is true for a locally finite collection, since then for any $x\in X$, to test whether $x\in\overline{\bigcup W_\gamma}$ you can restrict to a neighborhood of $x$ which only finitely many $W_\gamma$'s intersect and use the fact that a finite union of closed sets is closed. So without local finiteness, there is no reason to expect $\bigcup_\gamma\overline{W_\gamma}$ to be closed (after all, it is a possibly infinite union of closed sets!), and so there is no reason to expect $U$ to be open.