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  "body": "\n## Proof\n\nWe argue by contradiction.\n\n1. Suppose that $f$ is not [[uniformly continuous|0A3CK6DVVB]] on $[a, b]$. Then there is $\\varepsilon \u003e 0$ with the following property. For each $\\delta \u003e 0$ there are $x, y \\in [a, b]$ with $\\abs{x - y} \u003c \\delta$ and $\\abs{f(x) - f(y)} \\ge \\varepsilon$.\n2. For each $n \\in \\N$, apply step 1 with $\\delta = \\frac{1}{n+1}$. This gives $x_n, y_n \\in [a, b]$ with $\\abs{x_n - y_n} \u003c \\frac{1}{n+1}$ and $\\abs{f(x_n) - f(y_n)} \\ge \\varepsilon$.\n3. The sequence $(x_n)$ lies in $[a, b]$, so it is bounded. By the [[Bolzano–Weierstrass theorem|0KDJD249DT]], a [[subsequence|0R8J3ZQMJS]] $(x_{n_k})$ converges to some real $c$.\n4. Since $a \\le x_{n_k} \\le b$ for all $k$, [[limits preserve non-strict inequalities|0N32V0E3XJ]] gives $a \\le c \\le b$.\n5. The indices satisfy $n_0 \u003c n_1 \u003c \\cdots$, so induction gives $n_k \\ge k$. Hence $\\abs{x_{n_k} - y_{n_k}} \u003c \\frac{1}{n_k+1} \\le \\frac{1}{k+1}$.\n6. The function $f$ is [[continuous|0P27W9RW0G]] at $c$. By the [[epsilon-delta form of continuity|0W8YHHGCJG]], there is $\\delta \u003e 0$ with $\\abs{f(x) - f(c)} \u003c \\frac{\\varepsilon}{2}$ for all $x \\in [a, b]$ with $\\abs{x - c} \u003c \\delta$.\n7. By the [[epsilon-N form of limits|09G6KHS6GY]], there is $K_1$ with $\\abs{x_{n_k} - c} \u003c \\frac{\\delta}{2}$ for all $k \\ge K_1$. By [[this lemma|0WH916TJNY]], there is $K_2 \\ge 1$ with $\\frac{1}{K_2} \u003c \\frac{\\delta}{2}$.\n8. Let $k = \\max(K_1, K_2)$. Step 5 gives $\\abs{x_{n_k} - y_{n_k}} \u003c \\frac{1}{k+1} \u003c \\frac{1}{K_2} \u003c \\frac{\\delta}{2}$. By the [[triangle inequality|03HF5PMY5T]] and steps 5 and 7, $\\abs{y_{n_k} - c} \\le \\abs{y_{n_k} - x_{n_k}} + \\abs{x_{n_k} - c} \u003c \\frac{\\delta}{2} + \\frac{\\delta}{2} = \\delta$.\n9. Also $\\abs{x_{n_k} - c} \u003c \\delta$. So step 6 gives $\\abs{f(x_{n_k}) - f(c)} \u003c \\frac{\\varepsilon}{2}$ and $\\abs{f(y_{n_k}) - f(c)} \u003c \\frac{\\varepsilon}{2}$.\n10. By the triangle inequality, $\\abs{f(x_{n_k}) - f(y_{n_k})} \u003c \\varepsilon$. This contradicts step 2.\n11. So the assumption of step 1 is false, and $f$ is uniformly continuous on $[a, b]$.\n\n## Idea\n\nIf uniform continuity fails, there are pairs of points closer and closer together whose values stay $\\varepsilon$ apart. The Bolzano–Weierstrass theorem makes both points of these pairs converge to one point $c$ of the interval. Continuity at $c$ then forces the values together, which is a contradiction.\n",
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