[proofplan]
We reduce normality to the compact-compact separation result. Since $X$ is compact Hausdorff, every closed subset of $X$ is compact. The two disjoint closed sets $C_0, C_1$ are therefore compact, and the [Separation of Compact Sets in Hausdorff Spaces](/theorems/1026) (part 2) provides the required disjoint open sets.
[/proofplan]
[step:Verify that closed subsets of a compact Hausdorff space are compact]
Let $C_0, C_1 \subset X$ be disjoint closed subsets of the compact Hausdorff space $(X, \tau)$. Since $X$ is compact and $C_0$ is a closed subset of a compact space, $C_0$ is compact: if $\{W_\alpha\}_{\alpha \in A}$ is an open cover of $C_0$, then $\{W_\alpha\}_{\alpha \in A} \cup \{X \setminus C_0\}$ is an open cover of $X$, which admits a finite subcover by compactness of $X$, and removing the set $X \setminus C_0$ from this finite subcover yields a finite subcover of $C_0$. By the same argument, $C_1$ is compact.
[guided]
Let $C_0, C_1 \subset X$ be disjoint closed subsets of the compact Hausdorff space $(X, \tau)$. The first step is to observe that closed subsets of compact spaces are compact. We verify this explicitly for $C_0$ (the argument for $C_1$ is identical).
Let $\{W_\alpha\}_{\alpha \in A}$ be an arbitrary open cover of $C_0$. Since $C_0$ is closed, its complement $X \setminus C_0$ is open. The collection $\{W_\alpha\}_{\alpha \in A} \cup \{X \setminus C_0\}$ is then an open cover of $X$. Since $X$ is compact, this cover admits a finite subcover:
\begin{align*}
X \subset W_{\alpha_1} \cup \cdots \cup W_{\alpha_m} \cup (X \setminus C_0).
\end{align*}
In particular, $C_0 \subset W_{\alpha_1} \cup \cdots \cup W_{\alpha_m} \cup (X \setminus C_0)$. Since $C_0 \cap (X \setminus C_0) = \varnothing$, we have $C_0 \subset W_{\alpha_1} \cup \cdots \cup W_{\alpha_m}$. This is a finite subcover of $C_0$ from the original cover, so $C_0$ is compact. By the same argument applied to $C_1$, both sets are compact.
Why does this matter? Normality asks us to separate *closed* sets by open sets. In a general Hausdorff space, we only know how to separate *compact* sets (by the [Separation of Compact Sets in Hausdorff Spaces](/theorems/1026)). The compactness of $X$ bridges this gap: it ensures every closed set is automatically compact.
[/guided]
[/step]
[step:Apply compact-compact separation to obtain the required disjoint open sets]
Since $C_0$ and $C_1$ are disjoint compact subsets of the Hausdorff space $X$, the [Separation of Compact Sets in Hausdorff Spaces](/theorems/1026) (part 2) yields disjoint open sets $U_0, U_1 \in \tau$ with $C_0 \subset U_0$ and $C_1 \subset U_1$. This is the definition of normality.
[guided]
We now apply the [Separation of Compact Sets in Hausdorff Spaces](/theorems/1026), part (2). This theorem requires:
- $(X, \tau)$ is Hausdorff — given by hypothesis.
- $C_0$ and $C_1$ are compact — established in the previous step.
- $C_0 \cap C_1 = \varnothing$ — given by hypothesis.
All conditions are satisfied, so the theorem produces disjoint open sets $U_0, U_1 \in \tau$ with $C_0 \subset U_0$ and $C_1 \subset U_1$. Since $C_0$ and $C_1$ were arbitrary disjoint closed subsets of $X$, this establishes that $X$ is normal.
The proof reveals the structural reason that compact Hausdorff spaces are normal: the Hausdorff axiom provides separation of points, compactness upgrades this to separation of compact sets (via the finite extraction argument in Theorem 1026), and global compactness ensures every closed set is compact, completing the chain from point separation to closed-set separation.
[/guided]
[/step]
