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  "body": "\n## Proof\n\nWe argue by contradiction: in infinite dimension, Riesz's lemma gives unit vectors that stay far apart.\n\n1. Let $\\bar{B} = \\{x \\in X : \\|x\\| \\le 1\\}$ be compact, and suppose that $X$ is not [[finite-dimensional|018W3V3TCT]]. Then $X$ has no finite spanning set. In particular $X \\neq \\{0\\}$.\n2. We build unit vectors $x_1, x_2, \\ldots$ with $\\|x_m - x_n\\| \\ge \\tfrac{1}{2}$ for $m \\neq n$. Choose any $x_1 \\neq 0$ and replace it by $x_1 / \\|x_1\\|$.\n3. Given $x_1, \\ldots, x_n$, let $Y_n$ be their [[linear span|0HR573CHNM]]. It is a [[linear subspace|0EPTJA5QV4]], and $Y_n \\neq X$ because $X$ has no finite spanning set.\n4. $Y_n$ with the restricted norm is a finite-dimensional normed space. So it is complete, by [[finite-dimensional normed spaces are complete|0ABP2F2ZP7]].\n5. Its metric is the restriction of the metric of $X$. So $Y_n$ is closed in $X$, by [[complete subspaces are closed|0W41XG4W60]].\n6. By [[Riesz's lemma|097C7KCAT5]] with $\\alpha = \\tfrac{1}{2}$, there is $x_{n+1}$ with $\\|x_{n+1}\\| = 1$ and $\\|x_{n+1} - y\\| \\ge \\tfrac{1}{2}$ for all $y \\in Y_n$.\n7. For $m \u003e n$, we have $x_n \\in Y_{m-1}$. So $\\|x_m - x_n\\| \\ge \\tfrac{1}{2}$, by step 6 for $x_m$.\n8. All $x_n$ lie in $\\bar{B}$. As $\\bar{B}$ is compact in the metric space $X$, it is [[sequentially compact|0CARW59WD2]], by [[this result|0K8SGQSV7G]].\n9. So some subsequence $(x_{n_k})$ [[converges|06JYPJ4WHH]] to a point $z$. There is $k_0$ with $\\|x_{n_k} - z\\| \u003c \\tfrac{1}{4}$ for $k \\ge k_0$.\n10. Then $\\|x_{n_{k_0}} - x_{n_{k_0 + 1}}\\| \\le \\|x_{n_{k_0}} - z\\| + \\|z - x_{n_{k_0 + 1}}\\| \u003c \\tfrac{1}{2}$, with $n_{k_0} \\neq n_{k_0 + 1}$. This contradicts step 7.\n11. So $X$ is finite-dimensional.\n\n## Idea\n\nIn infinite dimension, each finite-dimensional subspace is a closed proper subspace. Riesz's lemma then gives a new unit vector far from all previous ones. The resulting sequence has no convergent subsequence, so the unit ball cannot be compact.\n",
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