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  "body": "\n## Proof\n\nWe prove the two directions separately. Throughout, the metric on $K$ is the [[subspace metric|0T2WYDTRP2]], so distances in $K$ are distances in $X$. We test compactness of $K$ with [[open covers by open subsets of X|0P6PMJB4MA]].\n\nFirst, let $K$ be compact. We show that $K$ is totally bounded.\n\n1. Fix a real $\\varepsilon \u003e 0$. The [[open balls|0DC3MPR7C9]] $B(y, \\varepsilon / 3)$, for $y \\in K$, are [[open|0B2305GNAZ]] in $X$ and cover $K$. So finitely many of them cover $K$.\n2. For $z, w \\in B(y, \\varepsilon / 3)$, the [[triangle inequality|01M6CW9VM6]] gives $d(z, w) \u003c 2 \\varepsilon / 3$. So each ball has [[diameter|0692CGQF00]] at most $2 \\varepsilon / 3 \u003c \\varepsilon$.\n3. Thus $K$ is covered by finitely many sets of diameter less than $\\varepsilon$, for each $\\varepsilon \u003e 0$. So $K$ is [[totally bounded|0WE5BBHGM6]].\n\nNext, let $K$ be compact. We show that $K$ is complete.\n\n4. Let $(x_n)$ be a [[Cauchy sequence|0MCK8K2D7Z]] of points of $K$. Suppose, for a contradiction, that it [[converges|06JYPJ4WHH]] to no point of $K$. Then $K$ is nonempty, since it contains $x_1$.\n5. Let $y \\in K$. Since $(x_n)$ does not converge to $y$, there is $\\varepsilon_y \u003e 0$ such that for each $N$ some $n \\ge N$ has $d(x_n, y) \\ge 2 \\varepsilon_y$.\n6. Since $(x_n)$ is Cauchy, there is $N_y$ with $d(x_m, x_n) \u003c \\varepsilon_y$ for all $m, n \\ge N_y$.\n7. By step 5, choose $m \\ge N_y$ with $d(x_m, y) \\ge 2 \\varepsilon_y$. For each $n \\ge N_y$, the triangle inequality gives $d(x_n, y) \\ge d(x_m, y) - d(x_m, x_n) \u003e \\varepsilon_y$.\n8. So the ball $B(y, \\varepsilon_y)$ contains no $x_n$ with $n \\ge N_y$.\n9. The balls $B(y, \\varepsilon_y)$, for $y \\in K$, are open and cover $K$. By compactness, finitely many cover $K$, say those with centres $y_1, \\ldots, y_p$.\n10. Let $N = \\max(N_{y_1}, \\ldots, N_{y_p})$. The point $x_N$ lies in $K$, so $x_N \\in B(y_j, \\varepsilon_{y_j})$ for some $j$. Since $N \\ge N_{y_j}$, this contradicts step 8.\n11. Hence each Cauchy sequence in $K$ converges to a point of $K$, and $K$ is [[complete|08EWT7J0SC]].\n\nConversely, let $K$ be complete and totally bounded. We argue by contradiction.\n\n12. Let $(U_i)_{i \\in I}$ be open subsets of $X$ with $K \\subseteq \\bigcup_{i \\in I} U_i$. Call a set $C \\subseteq K$ bad when no finite subfamily of $(U_i)$ covers $C$. Suppose that $K$ is bad.\n13. A bad set is nonempty, because the empty subfamily covers $\\varnothing$.\n14. Claim: if $C \\subseteq K$ is bad and $\\eta \u003e 0$, then some bad $C' \\subseteq C$ has $\\operatorname{diam}(C') \u003c \\eta$. Indeed, total boundedness gives sets $A_1, \\ldots, A_n$ of diameter less than $\\eta$ that cover $K$.\n15. Then $C = (C \\cap A_1) \\cup \\cdots \\cup (C \\cap A_n)$. If no $C \\cap A_k$ were bad, joining finitely many finite subfamilies would cover $C$. So some $C' = C \\cap A_k$ is bad, and $\\operatorname{diam}(C') \\le \\operatorname{diam}(A_k) \u003c \\eta$.\n16. Let $C_0 = K$. For $n \\ge 1$, apply step 14 to $C_{n-1}$ with $\\eta = 1/n$. We get bad sets $K \\supseteq C_1 \\supseteq C_2 \\supseteq \\cdots$ with $\\operatorname{diam}(C_n) \u003c 1/n$.\n17. By step 13, choose $x_n \\in C_n$ for each $n \\ge 1$. For $m \\ge n$, both $x_m$ and $x_n$ lie in $C_n$, so $d(x_m, x_n) \u003c 1/n$.\n18. Let $\\varepsilon \u003e 0$. Since [[the reals are Archimedean|0A3GWZ86AY]], there is $N$ with $N \\varepsilon \u003e 1$. Then $d(x_m, x_n) \u003c 1/N \u003c \\varepsilon$ for $m, n \\ge N$. So $(x_n)$ is Cauchy.\n19. Since $K$ is complete, $x_n \\to x$ for some $x \\in K$. Choose $i$ with $x \\in U_i$.\n20. The set $U_i$ is open in the [[metric topology|0A6YRNZB2R]]. So there is $r \u003e 0$ with $B(x, r) \\subseteq U_i$.\n21. Choose $n$ with $1/n \u003c r/2$ and $d(x_n, x) \u003c r/2$. This is possible by the Archimedean property and $x_n \\to x$.\n22. Let $z \\in C_n$. Then $d(z, x) \\le d(z, x_n) + d(x_n, x) \u003c 1/n + r/2 \u003c r$. So $C_n \\subseteq B(x, r) \\subseteq U_i$.\n23. Thus the single set $U_i$ covers $C_n$, and $C_n$ is not bad. This contradicts step 16.\n24. So some finite subfamily of $(U_i)$ covers $K$. By [[compactness via open covers|0P6PMJB4MA]], $K$ is compact.\n\n## Idea\n\nCompactness gives finite covers by small balls, which is total boundedness. A Cauchy sequence with no limit stays away from each point, so the balls witnessing this would need infinitely many members. For the converse, a cover with no finite subcover has a \"bad\" piece at every scale. The bad pieces shrink to a point, and the open set around that point swallows a whole bad piece.\n",
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