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  "body": "\n## Proof\n\nWe argue by contradiction. We build a nested sequence of closed balls with radii tending to $0$, and use completeness to get a point in all of them.\n\n1. Let $(X, d)$ be a [[complete metric space|08EWT7J0SC]]. Let $F_1, F_2, \\ldots$ be [[closed|0G5AV9Q4RR]] subsets of $X$ with $\\operatorname{int} F_n = \\emptyset$ for each $n$. Here $\\operatorname{int}$ is the [[interior|0C38S0KMJK]].\n2. Let $U_n = X \\setminus F_n$. Each $U_n$ is open, since $F_n$ is closed.\n3. Claim: each nonempty open set $V$ meets each $U_n$. Otherwise $V \\subseteq F_n$. The [[interior is the largest open subset|0HAZEFH137]], so $V \\subseteq \\operatorname{int} F_n = \\emptyset$, a contradiction.\n4. Let $W = \\operatorname{int} \\bigcup_n F_n$. By the same statement, $W$ is open and $W \\subseteq \\bigcup_n F_n$. Assume, for a contradiction, that $W \\ne \\emptyset$.\n5. By step 3, the set $W \\cap U_1$ is nonempty. It is open, since a [[topology|0CNQXVBHCX]] is closed under finite intersections. Pick $x_1$ in it.\n6. By the description of [[open sets of a metric space|0A26X79RTB]], there is $s \u003e 0$ with $B(x_1, s) \\subseteq W \\cap U_1$. Here $B$ is the [[open ball|0DC3MPR7C9]]. Let $r_1 = \\min(s / 2, 1/2)$.\n7. Let $\\bar{B}(x, r)$ be the [[closed ball|05DSV40EB9]]. Since $r_1 \u003c s$, we get $\\bar{B}(x_1, r_1) \\subseteq B(x_1, s) \\subseteq W \\cap U_1$.\n8. Suppose $x_n$ and $r_n \u003e 0$ are chosen. The set $B(x_n, r_n) \\cap U_{n+1}$ is open, since [[open balls are open|0B2305GNAZ]]. It is nonempty by step 3.\n9. As in steps 6 and 7, choose $x_{n+1}$ and $r_{n+1}$ with $0 \u003c r_{n+1} \\le 1/(n+1)$ and $\\bar{B}(x_{n+1}, r_{n+1}) \\subseteq B(x_n, r_n) \\cap U_{n+1}$.\n10. Write $\\bar{B}_n = \\bar{B}(x_n, r_n)$. By steps 7 and 9, $\\bar{B}_{n+1} \\subseteq \\bar{B}_n$, $\\bar{B}_n \\subseteq U_n$, $\\bar{B}_1 \\subseteq W$ and $r_n \\le 1/n$ for all $n$.\n11. By step 10, $\\bar{B}_m \\subseteq \\bar{B}_n$ for $m \\ge n$. So $x_m \\in \\bar{B}_n$ for all $m \\ge n$.\n12. For $m, p \\ge n$, the [[triangle inequality|01M6CW9VM6]] gives $d(x_m, x_p) \\le d(x_m, x_n) + d(x_n, x_p) \\le 2 r_n \\le 2/n$.\n13. Given $\\varepsilon \u003e 0$, choose $n \u003e 2/\\varepsilon$. Step 12 then shows that $(x_m)$ is a [[Cauchy sequence|0MCK8K2D7Z]].\n14. Since $X$ is complete, $(x_m)$ [[converges|06JYPJ4WHH]] to some $x \\in X$.\n15. Fix $n$. The tail $(x_m)_{m \\ge n}$ also converges to $x$, and all its terms lie in $\\bar{B}_n$ by step 11.\n16. [[Closed balls are closed|0YS40B38JD]], and [[closed sets contain their limits|09GQ3ZTQJ9]]. So $x \\in \\bar{B}_n$ for every $n$.\n17. By steps 10 and 16, $x \\in W$ and $x \\in U_n$ for every $n$. So $x \\in W$ but $x \\notin \\bigcup_n F_n$.\n18. This contradicts $W \\subseteq \\bigcup_n F_n$ from step 4. So $\\operatorname{int} \\bigcup_n F_n = \\emptyset$.\n19. This holds for every countable family of closed sets with empty interior. So $X$ is a [[Baire space|0D5KR33H7X]].\n\n## Idea\n\nA nonempty open set inside $\\bigcup_n F_n$ would leave room to shrink a ball away from $F_1$, then from $F_2$, and so on.\nThe centres of these nested balls form a Cauchy sequence, and completeness gives a limit. That limit lies in the open set but in no $F_n$.\n",
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