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  "body": "\n## Proof\n\nWe add the complement of $F$ to a cover of $F$ and use compactness of $K$.\n\n1. Let $X$ be a [[topological space|0CNQXVBHCX]], let $K \\subseteq X$ be [[compact|00R0X1WGPX]], and let $F \\subseteq K$ be [[closed|0G5AV9Q4RR]] in $X$.\n2. Let $(U_i)_{i \\in I}$ be [[open sets|01ZQQDCERP]] of $X$ with $F \\subseteq \\bigcup_{i \\in I} U_i$.\n3. The set $V = X \\setminus F$ is open, because $F$ is closed.\n4. Let $x \\in K$. If $x \\in F$, then $x$ lies in some $U_i$. If $x \\notin F$, then $x \\in V$. So the sets $U_i$ together with $V$ cover $K$.\n5. By [[compactness via open covers|0P6PMJB4MA]], finitely many of these sets cover $K$. So there are $i_1, \\ldots, i_n \\in I$ with $K \\subseteq U_{i_1} \\cup \\cdots \\cup U_{i_n} \\cup V$.\n6. Let $x \\in F$. Then $x \\in K$, so $x$ lies in one of the sets in step 5. But $x \\notin V$, so $x \\in U_{i_j}$ for some $j$.\n7. So $F \\subseteq U_{i_1} \\cup \\cdots \\cup U_{i_n}$. Each cover of $F$ by open sets of $X$ has a finite subcover.\n8. By compactness via open covers, $F$ is compact.\n\n## Idea\n\nThe complement of $F$ is open and covers everything in $K$ outside $F$.\nAdding it to a cover of $F$ gives a cover of $K$. A finite subcover of $K$, with the complement removed, still covers $F$.\n",
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