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  "body": "\n## Proof\n\nWe use an $\\varepsilon/3$ argument with the epsilon-delta form of continuity.\n\n1. Let $S \\subseteq \\R$, let each $f_n \\colon S \\to \\R$ be continuous, and let $f_n \\to f$ [[uniformly on $S$|08HJRTG5N7]].\n2. Fix $c \\in S$ and $\\varepsilon \u003e 0$. By the [[epsilon-delta form of continuity|0W8YHHGCJG]], it suffices to find $\\delta \u003e 0$ with $|f(x) - f(c)| \u003c \\varepsilon$ for all $x \\in S$ with $|x - c| \u003c \\delta$.\n3. By uniform convergence, there is $N$ with $|f_n(y) - f(y)| \u003c \\frac{\\varepsilon}{3}$ for all $n \\ge N$ and all $y \\in S$. In particular, this holds for $n = N$.\n4. The function $f_N$ is [[continuous at|0EEE2CZB2A]] $c$. By the epsilon-delta form, there is $\\delta \u003e 0$ with $|f_N(x) - f_N(c)| \u003c \\frac{\\varepsilon}{3}$ for all $x \\in S$ with $|x - c| \u003c \\delta$.\n5. Let $x \\in S$ with $|x - c| \u003c \\delta$. By the [[triangle inequality|03HF5PMY5T]],\n$$|f(x) - f(c)| \\le |f(x) - f_N(x)| + |f_N(x) - f_N(c)| + |f_N(c) - f(c)|.$$\n6. The first and third terms are less than $\\frac{\\varepsilon}{3}$ by step 3, with $y = x$ and $y = c$. The middle term is less than $\\frac{\\varepsilon}{3}$ by step 4.\n7. So $|f(x) - f(c)| \u003c \\varepsilon$. By step 2, $f$ is continuous at $c$.\n8. Since $c \\in S$ was arbitrary, $f$ is continuous on $S$.\n\n## Idea\n\nNear $c$, the limit $f$ is within $\\varepsilon/3$ of a single continuous function $f_N$, at every point at once. The function $f_N$ moves by less than $\\varepsilon/3$ near $c$. Three small errors add up to less than $\\varepsilon$. Uniformity is what lets one $N$ work at both $x$ and $c$.\n",
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