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  "body": "\n## Proof\n\nWe prove each implication. The second one goes through a finite net of balls and a Lebesgue number, both proved here for sequentially compact sets.\n\n1. Let $(X, d)$ be a [[metric space|063W5JMY85]] and let $K \\subseteq X$. Write $B(x, r)$ for the [[open ball|0DC3MPR7C9]]. [[Open balls are open|0B2305GNAZ]].\n2. Assume $K$ is [[compact|00R0X1WGPX]]. Let $(x_n)$ be a sequence in $K$. Suppose, for a contradiction, that no [[subsequence|0R8J3ZQMJS]] [[converges|06JYPJ4WHH]] to a point of $K$.\n3. Claim: each $y \\in K$ has a radius $r_y \u003e 0$ such that the set $I_y = \\set{n : x_n \\in B(y, r_y)}$ is finite.\n4. Proof of the claim: otherwise, for each $r \u003e 0$ there are infinitely many $n$ with $x_n \\in B(y, r)$. Then choose $n_1$ with $x_{n_1} \\in B(y, 1)$.\n5. Given $n_k$, choose $n_{k+1} \u003e n_k$ with $x_{n_{k+1}} \\in B(y, 1/(k+1))$. This is possible, since infinitely many indices qualify.\n6. Then $d(x_{n_k}, y) \u003c 1/k$ for all $k$, so $x_{n_k} \\to y \\in K$. This contradicts step 2 and proves the claim.\n7. The open balls $B(y, r_y)$ with $y \\in K$ cover $K$. By [[compactness via open covers|0P6PMJB4MA]], there are $y_1, \\ldots, y_m \\in K$ with $K \\subseteq B(y_1, r_{y_1}) \\cup \\cdots \\cup B(y_m, r_{y_m})$.\n8. Each $x_n$ lies in $K$, so in one of these balls. Hence every index $n$ lies in the finite set $I_{y_1} \\cup \\cdots \\cup I_{y_m}$.\n9. But there are infinitely many indices $n$. This contradiction shows that $K$ is [[sequentially compact|0CARW59WD2]].\n10. Now assume $K$ is sequentially compact. If $K = \\emptyset$, then $K$ is compact, since the empty subfamily covers it. So assume $K \\ne \\emptyset$.\n11. Claim A: for each $\\varepsilon \u003e 0$ there are finitely many points $a_1, \\ldots, a_p \\in K$ with $K \\subseteq B(a_1, \\varepsilon) \\cup \\cdots \\cup B(a_p, \\varepsilon)$.\n12. Proof of Claim A: suppose it fails for some $\\varepsilon$. Pick $a_1 \\in K$. Given $a_1, \\ldots, a_p$, their balls do not cover $K$, so pick $a_{p+1} \\in K$ outside all of them.\n13. Then $d(a_p, a_q) \\ge \\varepsilon$ for all $p \\ne q$. By sequential compactness, some subsequence $(a_{p_k})$ converges to a point $a \\in K$.\n14. For large $k$, $d(a_{p_k}, a) \u003c \\varepsilon / 2$. For two such distinct indices $k \u003c l$, the [[triangle inequality|01M6CW9VM6]] gives $d(a_{p_k}, a_{p_l}) \u003c \\varepsilon$. This contradicts step 13.\n15. Claim B: let $(U_i)_{i \\in I}$ be open sets of $X$ that cover $K$. Then there is $\\delta \u003e 0$ such that for each $x \\in K$, the ball $B(x, \\delta)$ lies in some $U_i$.\n16. Proof of Claim B: suppose not. Then for each $n \\ge 1$ there is $x_n \\in K$ such that $B(x_n, 1/n)$ lies in no $U_i$.\n17. By sequential compactness, a subsequence $(x_{n_k})$ converges to some $a \\in K$. Then $a \\in U_j$ for some $j$.\n18. By the description of [[open sets of a metric space|0A26X79RTB]], there is $r \u003e 0$ with $B(a, r) \\subseteq U_j$.\n19. Choose $k$ with $d(x_{n_k}, a) \u003c r / 2$ and $1/n_k \u003c r / 2$. This is possible, since $n_k \\ge k$ for the strictly increasing indices $n_k \\ge 1$.\n20. Let $z \\in B(x_{n_k}, 1/n_k)$. Then $d(z, a) \\le d(z, x_{n_k}) + d(x_{n_k}, a) \u003c r / 2 + r / 2 = r$. So $B(x_{n_k}, 1/n_k) \\subseteq B(a, r) \\subseteq U_j$.\n21. This contradicts the choice of $x_{n_k}$ in step 16, and proves Claim B.\n22. Now let $(U_i)_{i \\in I}$ be open sets of $X$ that cover $K$. Take $\\delta$ from Claim B, and apply Claim A with $\\varepsilon = \\delta$ to get $a_1, \\ldots, a_p \\in K$.\n23. For each $j$, Claim B gives an index $i_j$ with $B(a_j, \\delta) \\subseteq U_{i_j}$.\n24. Then $K \\subseteq B(a_1, \\delta) \\cup \\cdots \\cup B(a_p, \\delta) \\subseteq U_{i_1} \\cup \\cdots \\cup U_{i_p}$.\n25. So each cover of $K$ by open sets of $X$ has a finite subcover. By compactness via open covers, $K$ is compact.\n\n## Idea\n\nSuppose a sequence in a compact set had no convergent subsequence. Then each point has a ball that holds only finitely many terms, and finitely many such balls cover $K$, which is absurd.\nConversely, sequential compactness gives two uniform facts: $K$ is covered by finitely many $\\varepsilon$-balls, and one radius $\\delta$ fits each small ball into a set of the cover. Together they give a finite subcover.\n",
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