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  "body": "\n## Proof\n\nWe argue by contradiction.\n\n1. Let $a \\le b$ and let $f \\colon [a, b] \\to \\R$ be [[continuous|0P27W9RW0G]]. Suppose that $f$ is not bounded.\n2. Then for each natural number $n$, the number $n$ is not a bound. So there is $x_n \\in [a, b]$ with $\\abs{f(x_n)} \u003e n$.\n3. The sequence $(x_n)$ satisfies $a \\le x_n \\le b$, so it is [[bounded|05FQ3VGAR9]]. By the [[Bolzano–Weierstrass theorem|0KDJD249DT]], it has a [[subsequence|0R8J3ZQMJS]] $(x_{n_k})$ that converges to some real $c$.\n4. Since $a \\le x_{n_k} \\le b$ for all $k$, [[limits preserve non-strict inequalities|0N32V0E3XJ]], with constant sequences, gives $a \\le c \\le b$.\n5. The function $f$ is continuous, so it is [[sequentially continuous|0MVDQEAB60]] by the [[equivalence|0PAQM3H57X]]. Hence $f(x_{n_k}) \\to f(c)$.\n6. The convergent sequence $(f(x_{n_k}))$ is [[bounded|0S4JF619V7]]: there is $K$ with $\\abs{f(x_{n_k})} \\le K$ for all $k$.\n7. The indices satisfy $n_0 \u003c n_1 \u003c \\cdots$. By induction on $k$, $n_k \\ge k$ for all $k$.\n8. By the [[Archimedean property|0A74SX55QP]], there is a natural number $k$ with $k \u003e K$.\n9. For this $k$, steps 2 and 7 give $\\abs{f(x_{n_k})} \u003e n_k \\ge k \u003e K$. This contradicts step 6.\n10. So $f$ is bounded: there is $K$ with $\\abs{f(x)} \\le K$ for all $x \\in [a, b]$.\n\n## Idea\n\nIf $f$ were unbounded, we could pick points where $\\abs{f}$ exceeds $1, 2, 3, \\dots$. These points have a convergent subsequence, and its limit stays in the closed interval. Continuity at that limit would force the values to converge, hence stay bounded, which they do not.\n",
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