Rudin spaces are collectionwise normal

If $h,k\in X$ are distinct, choose a coordinate $n$ at which they differ, interchanging their names so $h(n)<k(n)$. Set $a(n)=h(n)$ and $a(i)=0$ for $i\mathrm{\ne }n$. Every point-coordinate is positive, so $a<k$. The clopen box $V=(a,k{]}_X$ contains $k$ and excludes $h$, since membership would require $h(n)>a(n)=h(n)$. Its open complement contains $h$ and is disjoint from $V$. Thus $X$ is Hausdorff.

Let $(F_j{)}_{j\in J}$ be an indexed discrete closed family in $X$ and put $C_j={\overline{F_j}}^Y$. By F3 and A1, each point of $Y$ has an open neighborhood meeting at most one $C_j$. Let $\mathcal{O}$ be the set of all open subsets of $Y$ with that property. It is an open cover by the preceding existence statement. F2 and A1 give a partition $\mathcal{V}$ of $Y$ into nonempty open boxes refining $\mathcal{O}$. Every cell meets at most one indexed $C_j$, since it is contained in a member of $\mathcal{O}$.

For each $j\in J$ define $V_j=\bigcup \{V\in \mathcal{V}:V\cap C_j\mathrm{\ne }\mathrm{\varnothing }\}$ and $U_j=V_j\cap X$. Each $V_j$ is open in $Y$, so $U_j$ is open in its subspace $X$. If $p\in F_j$, the unique partition cell through $p$ meets $C_j$ because $F_j\subseteq C_j$, and therefore $p\in U_j$. If $p\in V_j\cap V_l$, two partition cells assigned to $j$ and $l$ contain $p$. They are the same cell by disjointness of the partition. Step 1.2 then gives $j=l$. Thus the $V_j$, and consequently the $U_j$, are pairwise disjoint. For $F_j=\mathrm{\varnothing }$, also $C_j=\mathrm{\varnothing }$, so the displayed union gives $U_j=\mathrm{\varnothing }$. An empty index set produces the empty family.

Steps 1.1 and 2.1 give Hausdorffness and the stated collectionwise separation property. If $A,D$ are disjoint closed subsets of $X$, their two-member indexed family is discrete: at a point of $A$ the open complement of $D$ meets at most $A$, at a point of $D$ use the complement of $A$, and elsewhere the intersection of both complements meets neither. Applying step 2.1 gives disjoint open neighborhoods of $A,D$. Together with Hausdorffness this proves normality. QED.