§ Compact Hausdorff spaces are normal

Let $C, D$ be two disjoint closed subsets. We wish to exhibit disjoint opens $U, V$ which separate $C, D$. Formally, we want $C \subseteq U, D \subseteq V, U \cap V = \emptyset$. The crucial idea is to take all pairs of points in $C \times D$, and use Hausdorffness to find opens $\{ (U_{cd}, V_{cd}) : (c, d) \in C \times D \}$ such that $c \in U_{cd}, d \in V_{cd}, U_{cd} \cap V_{cd} = \emptyset$. which separate all pairs $c$ and $d$, and then to use compactness to escalate this into a real separating cover. Now that we have the pairs, for a fixed $c_0 \in C$, consider the cover $\cup_{d} V_{cd}$. This covers the set $D$, hence there is a finite subcover $D \subseteq V_{cD} \equiv \cup_{d_i} V_{c {d_i}}$. Now, we go back, and build the set $c \in U_{cD} \equiv \cap_{d_i} U_{c {d_i}}$. This is the intersection of a finite number of opens, and is hence open. So we now have two sets $U_{cD}$ and $V_{cD}$ which separate $c$ from $D$. We can build such a pair $U_{cD}, V_{cD}$ that separates each $c$ from all of $D$. Then, using compactness again, we find a finite subcover of sets $U_{c_i D}, V_{c_i D}$ such that the $U_{CD} \equiv \cup_{i=0}^n U_{c_i D}$ cover $C$, each of the $V_{c_i D}$ cover $D$ (so $V_{CD} \equiv \cap_{i=0}^n V_{c_i} D$ covers $D$). This gives us our final opens $U_{CD}$ and $V_{CD}$. that separate $C$ and $D$.