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  "body": "\n## Proof\n\nThe direct part uses general facts on metric spaces. For the converse, we show that $K$ is sequentially compact, coordinate by coordinate.\n\n1. Let $n \\ge 1$, and let $d_2$ be the [[Euclidean metric|0E1MMCAY5B]] on $\\R^n$. Let $K \\subseteq \\R^n$.\n2. Assume $K$ is [[compact|00R0X1WGPX]]. Since [[compact subsets of a metric space are bounded|0R7BTEWCJ2]], $K$ is [[bounded|0P953MG05E]].\n3. [[Metric spaces are Hausdorff|0P9DCDES73]], and [[compact subsets of a Hausdorff space are closed|0MYFJVMN7M]]. So $K$ is [[closed|0G5AV9Q4RR]].\n4. Now assume $K$ is closed and bounded. If $K = \\emptyset$, it is compact. So assume $K \\ne \\emptyset$, fix $a \\in K$, and let $M$ be real with $d_2(x, y) \\le M$ for $x, y \\in K$.\n5. For $x \\in K$ and each coordinate $i$, $\\abs{x_i - a_i} \\le d_2(x, a) \\le M$, since one term of a sum of squares is at most the sum. So $\\abs{x_i} \\le \\abs{a_i} + M$.\n6. Let $(p^{(m)})_{m \\ge 1}$ be a sequence in $K$. By step 5, each coordinate sequence $(p^{(m)}_i)_m$ is a [[bounded real sequence|05FQ3VGAR9]], also along any subsequence.\n7. By the [[Bolzano–Weierstrass theorem|0KDJD249DT]], the first coordinates have a convergent [[subsequence|0R8J3ZQMJS]]. Pass to it.\n8. Repeat for coordinates $2, \\ldots, n$: at step $j$, apply Bolzano–Weierstrass to the $j$-th coordinates along the current subsequence, and pass to a further subsequence.\n9. A subsequence of a subsequence is a subsequence. [[Subsequences of a convergent sequence converge to the same limit|0VHAVXW08V]], so coordinates that converged earlier still converge.\n10. After $n$ steps we get indices $m_1 \u003c m_2 \u003c \\cdots$ and reals $c_1, \\ldots, c_n$ with $p^{(m_k)}_i \\to c_i$ for each $i$. Let $c = (c_1, \\ldots, c_n)$.\n11. For reals $t_1, \\ldots, t_n$, expanding the square gives $(\\sum_i \\abs{t_i})^2 \\ge \\sum_i t_i^2$, since the cross terms are $\\ge 0$. So $\\sqrt{\\sum_i t_i^2} \\le \\sum_i \\abs{t_i}$.\n12. By step 11, $d_2(p^{(m_k)}, c) \\le \\sum_{i=1}^n \\abs{p^{(m_k)}_i - c_i}$.\n13. Let $\\varepsilon \u003e 0$. For each $i$, there is $L_i$ with $\\abs{p^{(m_k)}_i - c_i} \u003c \\varepsilon / n$ for $k \\ge L_i$. For $k \\ge \\max_i L_i$, step 12 gives $d_2(p^{(m_k)}, c) \u003c \\varepsilon$.\n14. So $p^{(m_k)} \\to c$ in $(\\R^n, d_2)$, in the sense of [[convergence|06JYPJ4WHH]].\n15. All $p^{(m_k)}$ lie in $K$, and $K$ is closed. Since [[closed sets contain their limits|09GQ3ZTQJ9]], $c \\in K$.\n16. So every sequence in $K$ has a subsequence that converges to a point of $K$. Thus $K$ is [[sequentially compact|0CARW59WD2]].\n17. Since [[compactness and sequential compactness agree in metric spaces|0K8SGQSV7G]], $K$ is compact.\n\n## Idea\n\nCompact sets in any metric space are closed and bounded, so the work is in the converse.\nA bounded sequence in $\\R^n$ has bounded coordinates. Bolzano–Weierstrass, applied once per coordinate, gives a convergent subsequence, and closedness keeps its limit in $K$.\n",
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