Proof of the uniform limit theorem by an epsilon over three argument

Proof

We use an $\varepsilon/3$ argument with the epsilon-delta form of continuity.

1. Let $S \subseteq \R$, let each $f_n \colon S \to \R$ be continuous, and let $f_n \to f$ uniformly on $S$.
2. Fix $c \in S$ and $\varepsilon > 0$. By the epsilon-delta form of continuity, it suffices to find $\delta > 0$ with $|f(x) - f(c)| < \varepsilon$ for all $x \in S$ with $|x - c| < \delta$.
3. By uniform convergence, there is $N$ with $|f_n(y) - f(y)| < \frac{\varepsilon}{3}$ for all $n \ge N$ and all $y \in S$. In particular, this holds for $n = N$.
4. The function $f_N$ is continuous at $c$. By the epsilon-delta form, there is $\delta > 0$ with $|f_N(x) - f_N(c)| < \frac{\varepsilon}{3}$ for all $x \in S$ with $|x - c| < \delta$.
5. Let $x \in S$ with $|x - c| < \delta$. By the triangle inequality, $$|f(x) - f(c)| \le |f(x) - f_N(x)| + |f_N(x) - f_N(c)| + |f_N(c) - f(c)|.$$
6. 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.
7. So $|f(x) - f(c)| < \varepsilon$. By step 2, $f$ is continuous at $c$.
8. Since $c \in S$ was arbitrary, $f$ is continuous on $S$.