= Solution
Fix $\epsilon>0$. For each $N$, the set $Q_N=\bigcap_{n,m\geq N}\{x:|f_n(x)-f_m(x)|\leq\epsilon\}$ is <closed>, because each $f_n-f_m$ is <continuous>. At every $x$, <pointwise convergence> makes $(f_n(x))$ a <Cauchy sequence>, so $\mathbb R=\bigcup_{N\geq1}Q_N$. The <Baire category theorem> supplies an $N_0$ for which $Q_{N_0}$ contains a nonempty <open interval> $I$. For $x\in I$ and $n\geq N_0$, pass $m\to\infty$ in $|f_n(x)-f_m(x)|\leq\epsilon$ to obtain
$$
\boxed{|f_n(x)-f(x)|\leq\epsilon\quad(x\in I,\ n\geq N_0).}
$$
This <local uniform Cauchy control for pointwise convergent continuous functions> allows both $I$ and $N_0$ to depend on $\epsilon$.
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