Can anyone help me verify a piece of a proof which shows that normality implies that any point-finite open covering can be shrunk? Can anyone help me verify a piece of a proof which shows that normality implies that any point-finite open covering can be shrunk?
To be more specific, the proof is in Willard's General Topology: link to Google books.
I cannot verify that the $F_{\alpha}$ which they define inductively is contained in the $U_{\alpha}$.  In fact, by drawing a picture, I think I can find a counterexample.  However, I think Willard knows what he's doing so there's likely something I don't see.
 A: You have $$F_\alpha = X \setminus \left(\bigcup\limits_{\beta<\alpha}V_\beta \cup \bigcup\limits_{\gamma>\alpha}U_\gamma \right),$$ and your induction hypothesis is that for each $\beta < \alpha$, $F_\beta \subseteq V_\beta \subseteq\operatorname{cl}V_\beta \subseteq U_\beta$, $F_\beta$ is closed, and $V_\beta$ is open. Certainly $F_\alpha$ is closed, since it’s the complement of an open set. Now note that $F_\alpha \subseteq U_\alpha$ iff $$U_\alpha \cup \bigcup\limits_{\beta<\alpha}V_\beta \cup \bigcup\limits_{\gamma>\alpha}U_\gamma = X,$$ i.e., iff $\{V_\beta:\beta < \alpha\} \cup \{U_\gamma:\gamma \ge \alpha\}$ is an open cover of $X$.
Suppose that $\{V_\beta:\beta < \alpha\} \cup \{U_\gamma:\gamma \ge \alpha\}$ doesn’t cover $X$, and let $$H = X \setminus \left(\bigcup\limits_{\beta<\alpha}V_\beta \cup \bigcup\limits_{\gamma\ge\alpha}U_\gamma\right).$$ Let $x$ be any point of $H$. $\mathscr{U}$ is point-finite, so $x$ is in only finitely many members of $\mathscr{U}$, and we can let $\mu$ be the largest index such that $x \in U_\mu$. This clearly implies that $x \notin \bigcup\limits_{\gamma>\mu}U_\gamma$. We also know that $x \notin \bigcup\limits_{\beta<\alpha}V_\beta$, since $x\in H$, and since $\mu < \alpha$, $x \notin \bigcup\limits_{\beta<\mu}V_\beta$. Thus, $$x \notin \bigcup\limits_{\beta<\mu}V_\beta \cup \bigcup\limits_{\gamma>\mu}U_\gamma.$$
But by definition $$F_\mu = X \setminus \left(\bigcup\limits_{\beta<\mu}V_\beta \cup \bigcup\limits_{\gamma>\mu}U_\gamma \right),$$ so $x \in F_\mu \subseteq V_\mu \subseteq X \setminus H$, which is a contradiction: $H$ is disjoint from $V_\mu$. Therefore $H=\varnothing$, $\{V_\beta:\beta < \alpha\} \cup \{U_\gamma:\gamma \ge \alpha\}$ is an open cover of $X$, and $F_\alpha \subseteq U_\alpha$.
Added: It may help your intuition to think of it this way. You’re trying to shrink the members of $\mathscr{U}$ one at a time without losing coverage of the whole space. At stage $\alpha$ the set $F_\alpha$ is everything that isn’t covered by the $V_\beta$ with $\beta<\alpha$ or by the $U_\gamma$ with $\gamma>\alpha$, so it's the part of $X$ that at this point in the shrinking process is covered only by $U_\alpha$. We then choose $V_\alpha$ to cover it while still staying well inside $U_\alpha$, so that we still have a cover of $X$. The reason that this isn’t a rigorous argument is that I’ve simply assumed here that $\{V_\beta:\beta<\alpha\} \cup \{U_\gamma:\gamma\ge\alpha\}$ covers $X$. Showing this is the actual work that needs to be done and what the argument above actually does.
