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  "body": "\n## Proof\n\nWe first separate a point from a compact set, and then a compact set from a compact set.\n\n1. Let $A, B \\subseteq X$ be disjoint [[closed sets|0G5AV9Q4RR]]. The [[subspace topology|0D5YHQGDJY]] on $X$ as a subset of itself is its own topology, so $X$ is a [[compact set|00R0X1WGPX]] in $X$.\n2. By [[closed subsets of compact sets are compact|066HW6D9A7]], $A$ and $B$ are compact.\n3. Claim: if $C \\subseteq X$ is compact and $x \\notin C$, there are disjoint open $U \\ni x$ and $V \\supseteq C$. If $C = \\varnothing$, take $U = X$ and $V = \\varnothing$.\n4. Let $C \\ne \\varnothing$. For each $c \\in C$, we have $c \\neq x$. Since $X$ is [[Hausdorff|0MXB4CKHFW]], choose disjoint open sets $U_c \\ni x$ and $V_c \\ni c$.\n5. The open sets $V_c$, for $c \\in C$, cover $C$. By [[compactness via open covers|0P6PMJB4MA]], there are $c_1, \\ldots, c_n \\in C$ with $C \\subseteq V_{c_1} \\cup \\cdots \\cup V_{c_n}$.\n6. Let $U = U_{c_1} \\cap \\cdots \\cap U_{c_n}$ and $V = V_{c_1} \\cup \\cdots \\cup V_{c_n}$. Both are open by the axioms of a [[topology|0CNQXVBHCX]]. Also $x \\in U$ and $C \\subseteq V$.\n7. If $z \\in U \\cap V$, then $z \\in V_{c_k}$ for some $k$ and $z \\in U_{c_k}$. This contradicts $U_{c_k} \\cap V_{c_k} = \\varnothing$. So $U \\cap V = \\varnothing$, and the claim holds.\n8. Now separate $A$ and $B$. If $A = \\varnothing$, take $U = \\varnothing$ and $V = X$. Let $A \\ne \\varnothing$.\n9. For each $a \\in A$, we have $a \\notin B$. Step 3 with $C = B$ gives disjoint open sets $U_a \\ni a$ and $V_a \\supseteq B$.\n10. The open sets $U_a$, for $a \\in A$, cover $A$. By compactness of $A$, there are $a_1, \\ldots, a_m \\in A$ with $A \\subseteq U_{a_1} \\cup \\cdots \\cup U_{a_m}$.\n11. Let $U = U_{a_1} \\cup \\cdots \\cup U_{a_m}$ and $V = V_{a_1} \\cap \\cdots \\cap V_{a_m}$. Both are open, $A \\subseteq U$ and $B \\subseteq V$.\n12. If $z \\in U \\cap V$, then $z \\in U_{a_k}$ for some $k$ and $z \\in V_{a_k}$, which contradicts $U_{a_k} \\cap V_{a_k} = \\varnothing$. So $U \\cap V = \\varnothing$.\n13. Hence $X$ is [[normal|0AJ646N339]].\n\n## Idea\n\nThe Hausdorff property separates two points. Compactness upgrades this, one finite cover at a time: first a point from a compact set, then a compact set from a compact set. Closed sets of a compact space are compact, so this separates any two disjoint closed sets.\n",
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