 Solution

The sequence $(f_n)$ converges by uniform convergence to $f$ on $E$ ifThe index $M$ must work simultaneously for every $x\in E$.

Given $\epsilon >0$, choose $m\ge N$ so large that $|f_m(x)-f(x)|<\epsilon /3$ for every $x\in E$. The continuity of this one fixed $f_m$ at $x_0$ gives $\delta >0$ such that $x\in E$ and $|x-x_0|<\delta$ imply $|f_m(x)-f_m(x_0)|<\epsilon /3$. The triangle inequality then givesThus $f$ is continuous at $x_0$, with continuity understood relative to $E$. This is the pointwise form of the uniform limit theorem.