I was reading Munkres' Topology book and I came across this Shrinking Lemma: If {$U_1, ..., U_n$} is an open cover of a normal space $X$, then there is an open cover {$V_1, ..., V_n$} such that the closure $\overline{V_i} \subset U_i$ for each $i = 1,...,n$. The proof goes like this:
$A = X - (U_2\cup ...\cup U_n)$ is closed and since {$U_i$} covers $X$, $A \subset U_1$. Since $X$ is normal, there is an open set $V_1$ such that $A \subset V_1$ and $\overline{V_1} \subset U_1$. The collection {$V_1, U_2, ..., U_n$} covers X. In general, given open sets $V_1, ..., V_{k-1}$ such that {$V_1, ...,V_{k-1}, U_k, U_{k+1},...,U_n$} covers $X$, let $A = X - (V_1\cup ...\cup V_{k-1}) - (U_{k+1}\cup ...\cup U_n)$. $A$ is closed so there exists an open set $V_k$ such that $A \subset V_k$ and $\overline{V_k} \subset U_k$. {$V_1, ...,V_k, U_{k+1},...,U_n$} is an open cover of $X$. At the $n$th step of the induction, our result is proved.
This obviously generalizes to a countably infinite open cover. What about an uncountable open cover (of nonempty sets)? Does the proof below work?
Suppose $X$ is a normal space and {$U_\alpha$}$_{\alpha \in J}$ is an open cover of $X$ where $J$ is an uncountable index set and each $U_\alpha$ is nonempty. There is an order relation on $J$ that is a well-ordering by the well-ordering theorem. Let $\alpha_0$ be the smallest element of $J$. The set $A = X - \bigcup\limits_{\alpha \in J - \{\alpha_0\}} U_\alpha$ is closed and is a subset of $U_{\alpha_0}$ so there is an open set $V_{\alpha_0}$ such that $A \subset V_{\alpha_0} \subset \overline{V_{\alpha_0}} \subset U_{\alpha_0}$. Then {$U_\alpha\space\vert\space\alpha \in J - \{\alpha_0\}$}$\cup${$V_{\alpha_0}$} is an open cover of $X$. Let $\mathcal{C}$ = {open sets $V\space\vert\space\overline{V} \subset U_\alpha$ for some $\alpha \in J$} and $\mathcal{F}$ be the set of functions mapping a section of $J$ into $\mathcal{C}$. For all $\alpha \in J$, let $B_\alpha = \{V \in \mathcal{C}\space\vert\space \overline{V} \subset U_\alpha$ and $X - \bigcup\limits_{\lambda < \alpha}f(\lambda) - \bigcup\limits_{\lambda > \alpha}U_\lambda \subset V$ for some $f \in \mathcal{F}$ with domain $S_\alpha$}. If it's unclear, a section of $J$ is just $S_\alpha =$ {$\lambda \in J\space \vert\space \lambda < \alpha$} for some $\alpha \in J$. Let $\mathcal{B}$ = {$B_\alpha\space\vert\space\alpha \in J$}. By the axiom of choice, there is a choice function $c : \mathcal{B} \longrightarrow \bigcup\limits_{B \in \mathcal{B}}B = \mathcal{C}$ such that $c(B) \in B\space\forall\space B \in \mathcal{B}$. Define $\rho : \mathcal{F} \longrightarrow \mathcal{C}$ by $\rho(f) = c(B_\alpha)$ where $S_\alpha$ is the domain of $f$. Then by the general principle of recursive definition, there is a unique function $h : J \longrightarrow\mathcal{C}$ such that $h(\alpha) = \rho(h\vert S_\alpha)$. So for each $\alpha \in J$, we have an open set $V_\alpha = h(\alpha) = \rho(h\vert S_\alpha) = c(B_\alpha)$ such that $\overline{V_\alpha} \subset U_\alpha$ and $X - \bigcup\limits_{\lambda < \alpha}h(\lambda) - \bigcup\limits_{\lambda > \alpha}U_\lambda \subset V_\alpha$. The set {$V_\alpha$} is an open cover of $X$.
