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  "body": "\n## Proof\n\nWe separate the two closed sets by comparing the distances to them.\n\n1. Since $X$ is [[metrizable|09KDHB82SG]], choose a [[metric|063W5JMY85]] $d$ on $X$ whose [[metric topology|0A6YRNZB2R]] is the topology of $X$. Let $A, B \\subseteq X$ be disjoint [[closed sets|0G5AV9Q4RR]].\n2. If $A = \\varnothing$, take $U = \\varnothing$ and $V = X$. If $B = \\varnothing$, take $U = X$ and $V = \\varnothing$. From now on let $A$ and $B$ be nonempty.\n3. For nonempty $C \\subseteq X$, write $d(x, C)$ for the [[distance from x to C|07TXYYVYZR]]. The metric axioms give $d \\ge 0$, so $d(x, C) \\ge 0$.\n4. Claim: $d(y, C) \\le d(y, x) + d(x, C)$ for all $x, y \\in X$. For each $c \\in C$, the [[triangle inequality|01M6CW9VM6]] gives $d(y, C) \\le d(y, c) \\le d(y, x) + d(x, c)$.\n5. So $d(y, C) - d(y, x)$ is a lower bound of $\\set{d(x, c) : c \\in C}$. By the definition of the [[infimum|04KFTA34C5]], it is at most $d(x, C)$. This proves the claim.\n6. Claim: if $C$ is closed and $x \\notin C$, then $d(x, C) \u003e 0$. The set $X \\setminus C$ is open and contains $x$, so some [[open ball|0DC3MPR7C9]] $B(x, r)$ with $r \u003e 0$ misses $C$.\n7. Then $d(x, c) \\ge r$ for each $c \\in C$, so $d(x, C) \\ge r \u003e 0$.\n8. Let $U = \\set{x : d(x, A) \u003c d(x, B)}$ and $V = \\set{x : d(x, B) \u003c d(x, A)}$. No point satisfies both inequalities, so $U \\cap V = \\varnothing$.\n9. Let $a \\in A$. Then $d(a, A) = 0$, since $d(a, a) = 0$. Since $a \\notin B$ and $B$ is closed, step 6 gives $d(a, B) \u003e 0$. So $a \\in U$, and $A \\subseteq U$.\n10. In the same way, $B \\subseteq V$.\n11. We show that $U$ is open. Let $x \\in U$ and $\\eta = d(x, B) - d(x, A) \u003e 0$. Let $y \\in B(x, \\eta / 2)$.\n12. By step 4, $d(y, A) \\le d(x, A) + d(x, y)$. Also by step 4, $d(x, B) \\le d(x, y) + d(y, B)$, so $d(y, B) \\ge d(x, B) - d(x, y)$.\n13. Subtracting, $d(y, B) - d(y, A) \\ge \\eta - 2 d(x, y) \u003e 0$. So $y \\in U$.\n14. Thus $B(x, \\eta / 2) \\subseteq U$ for each $x \\in U$, and $U$ is open in the metric topology. In the same way, $V$ is open.\n15. So $U$ and $V$ are disjoint open sets with $A \\subseteq U$ and $B \\subseteq V$. Hence $X$ is [[normal|0AJ646N339]].\n\n## Idea\n\nPut each point on the side of the closed set it is nearer to. Points of $A$ are at distance $0$ from $A$ and at a positive distance from the closed set $B$, so they land on the side of $A$. Distances to a set change by at most the distance moved, so a strict inequality survives small moves, and both sides are open.\n",
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