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  "body": "\n## Proof\n\nWe pass to complements and use [[compactness via open covers|0P6PMJB4MA]]. We read \"finitely many indices\" as allowing none. For no index, the intersection $K \\cap F_{i_1} \\cap \\cdots \\cap F_{i_n}$ is $K$ itself.\n\n1. Let $X$ be a [[topological space|0CNQXVBHCX]], let $K \\subseteq X$, and let $(F_i)_{i \\in I}$ be [[closed subsets|0G5AV9Q4RR]] of $X$. Let $U_i = X \\setminus F_i$, an [[open set|01ZQQDCERP]].\n2. Claim: for each $J \\subseteq I$, $K \\subseteq \\bigcup_{j \\in J} U_j$ holds if and only if $K \\cap \\bigcap_{j \\in J} F_j = \\emptyset$. For $J = \\emptyset$, the empty union is $\\emptyset$ and $K \\cap \\bigcap_{j \\in \\emptyset} F_j$ means $K$.\n3. Proof of the claim: a point $x \\in K$ lies outside $\\bigcap_{j \\in J} F_j$ if and only if $x \\notin F_j$ for some $j \\in J$. This means $x \\in U_j$ for some $j \\in J$. So each point of $K$ lies in $\\bigcup_{j \\in J} U_j$ if and only if no point of $K$ lies in $\\bigcap_{j \\in J} F_j$.\n4. Assume $K$ is [[compact|00R0X1WGPX]]. Let $(F_i)$ satisfy $K \\cap F_{i_1} \\cap \\cdots \\cap F_{i_n} \\ne \\emptyset$ for all finitely many indices.\n5. Suppose $K \\cap \\bigcap_{i \\in I} F_i = \\emptyset$. By step 2 with $J = I$, the open sets $U_i$ cover $K$.\n6. By compactness via open covers, there are finitely many indices $i_1, \\ldots, i_n$ with $K \\subseteq U_{i_1} \\cup \\cdots \\cup U_{i_n}$.\n7. By step 2 with $J = \\set{i_1, \\ldots, i_n}$, this gives $K \\cap F_{i_1} \\cap \\cdots \\cap F_{i_n} = \\emptyset$. This contradicts step 4.\n8. So $K \\cap \\bigcap_{i \\in I} F_i \\ne \\emptyset$, and $K$ has the property.\n9. Now assume $K$ has the property. Let $(U_i)_{i \\in I}$ be open sets of $X$ that cover $K$. Let $F_i = X \\setminus U_i$, a closed set.\n10. Suppose no finite subfamily of $(U_i)$ covers $K$. By step 2, $K \\cap F_{i_1} \\cap \\cdots \\cap F_{i_n} \\ne \\emptyset$ for all finitely many indices.\n11. By the property, $K \\cap \\bigcap_{i \\in I} F_i \\ne \\emptyset$. By step 2 with $J = I$, the sets $U_i$ do not cover $K$. This contradicts step 9.\n12. So some finite subfamily of $(U_i)$ covers $K$. By compactness via open covers, $K$ is compact.\n\n## Idea\n\nClosed sets with an empty intersection inside $K$ are the same as open sets, their complements, that cover $K$.\nSo \"every open cover has a finite subcover\" turns into \"if all finite intersections meet $K$, the whole intersection meets $K$\".\n",
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