Uniform limit theorem for continuity

Uniform limit theorem for continuity: Let $(X,d)$ be a metric space and let $(f_n)$ be a sequence of continuous functions $f_n:X\to\mathbb{R}$ (or into any metric space). If $f_n\to f$ uniformly on $X$, then $f$ is continuous on $X$.

Uniform convergence is strong enough to pass continuity through the limit. This is a fundamental reason uniform convergence is preferred over pointwise convergence in analysis.

Proof sketch: Fix $x_0\in X$ and $\varepsilon>0$. Choose $N$ such that $\sup_{x\in X} d(f_n(x),f(x))<\varepsilon/3$ for all $n\ge N$. By continuity of $f_N$ at $x_0$, choose $\delta>0$ so that $d(x,x_0)<\delta$ implies $d(f_N(x),f_N(x_0))<\varepsilon/3$. Then $d(f(x),f(x_0))\le d(f(x),f_N(x))+d(f_N(x),f_N(x_0))+d(f_N(x_0),f(x_0))<\varepsilon.$