Normal space and discrete family If $X$ is a normal space and $\{ F_\alpha : \alpha \in A \}$ is a countable discrete family of closed sets in $X$, then how to prove it that we can find  a discrete family $\{O_\alpha : \alpha \in A \} $ of open sets in $X$ such that $F_\alpha \subset O_\alpha$ for every $\alpha \in A$.
 A: Let $F_1, F_2, \ldots$ be a countable discrete family of closed sets in $X$. (Recall that a family $A_a, a \in A$ of subsets of $X$ is discrete iff every point of $X$ has a neighbourhood $O$ such that $\{a \in A: A_a \cap O \neq \emptyset\}$ has size at most one; this implies mutual disjointness but is stronger than that.)
Note that for any $I \subset \mathbb{N}$, the set $\cup_{i \in I} F_i$ is also closed in $X$, by the discreteness of the family. (Check this is this is new to you!).
So we can find for each $i$ find $U_i, V_i$ open and disjoint in $X$ such that $F_i \subseteq U_i$ and $\cup_{j \neq i} F_i \subseteq V_i$, by applying normality of $X$ and the above fact for each $i$. Define $O_1 = U_1$ and $O_i = U_i \cap V_1 \cap \ldots \cap V_{i-1}$ for $i > 1$. If $i \neq j$ (say $i < j$) then $O_i \cap O_j \subseteq V_i \cap U_i = \emptyset$, and $F_i \subseteq O_i$ for all $i$ (why?). 
This gives us a pairwise disjoint family of open sets $O_1,O_2,\ldots,$ such that $F_i \subseteq O_i$ for all $i$. We need to turn it into a discrete family.
Define $H = X \setminus (\cup_i O_i)$, $K = \cup_i F_i$. $H$ and $K$ are disjoint subsets of $X$. If $H = \emptyset$, this means the $O_i$ covered $X$ (disjointly) so they are closed and open in $X$ (why?) and so they already form a discrete family (why?).
So assume $H \neq \emptyset$. Separate $H$ and $K$ by disjoint open sets (using the normality of $X$ again), say $H \subseteq O_H$ and $K \subseteq O_K$. Define $O'_i = O_i \cap O_K$. We claim that these form a discrete open family. It's already clear that still $F_i \subseteq O'_i$, as $F_i \subseteq K \subseteq O_K$.
To see discreteness: Let $x \in X$. First suppose $x \notin H$, so $x \in O_i$ for some $i$. Then $O_i$ itself is the required neighbourhood of $x$ that intersects only one element of the $O'_i$ family. On the other hand, if $x \in H$, then $O_H$ misses $O_K$ and thus all $O'_i$ and again, we have a neighbourhood that intersects at most one (here none) of the sets $O'_i$. 
Note that we use the countability in the definition of the $O_i$: we have only finitely many predecessors for every $i$, and a finite intersection of open sets is still open. The last trick, to go from disjoint open neighbourhoods to a discrete family of open neighbourhoods is quite general. The countability is quite essential: if we can separate any sized discrete family of closed sets by discrete (or pairwise disjoint) open neighbourhoods, we have the notion of a "collectionwise normal space", which is stronger than being normal (it implies normal, as a collection of two disjoint closed sets is discrete). For a famous example of this see this blog post on Bing's G space.    
