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  "body": "\n## Proof\n\nWe cover $K$ by open balls of growing radius around one point.\n\n1. Let $(X, d)$ be a [[metric space|063W5JMY85]] and let $K$ be a [[compact|00R0X1WGPX]] subset of $X$.\n2. If $K = \\emptyset$, then $K$ is [[bounded|0P953MG05E]], since the condition on pairs of points is empty. So assume $K \\ne \\emptyset$ and fix $a \\in K$.\n3. For each integer $n \\ge 1$, let $B(a, n)$ be the [[open ball|0DC3MPR7C9]]. It is open, since [[open balls are open|0B2305GNAZ]].\n4. Let $x \\in K$. Choose an integer $n \u003e d(a, x)$. Then $x \\in B(a, n)$. So the balls $B(a, n)$ with $n \\ge 1$ cover $K$.\n5. By [[compactness via open covers|0P6PMJB4MA]], there are integers $n_1, \\ldots, n_p \\ge 1$ with $K \\subseteq B(a, n_1) \\cup \\cdots \\cup B(a, n_p)$. Since $K \\ne \\emptyset$, $p \\ge 1$.\n6. Let $N = \\max(n_1, \\ldots, n_p)$. Each $B(a, n_j)$ lies in $B(a, N)$, so $K \\subseteq B(a, N)$.\n7. Let $x, y \\in K$. By the [[triangle inequality|01M6CW9VM6]], $d(x, y) \\le d(x, a) + d(a, y) \u003c 2N$.\n8. So $d(x, y) \\le 2N$ for all $x, y \\in K$, and $K$ is bounded.\n\n## Idea\n\nThe balls of radius $1, 2, 3, \\ldots$ around a point exhaust the whole space.\nCompactness keeps only finitely many of them, and the largest one already holds all of $K$.\n",
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