= Solution
The <uniform convergence> condition on a set $X$ is
$$
\boxed{\forall\epsilon>0\ \exists N\ \forall n\ge N\ \forall x\in X:\quad |f_n(x)-f(x)|<\epsilon.}
$$
The index $N$ is independent of $x$; this is stronger than <pointwise convergence>.
Fix $x_0\in\mathbb R$ and $\epsilon>0$. Choose one $N$ such that $|f_N(x)-f(x)|<\epsilon/3$ for every $x$. The <continuity> of $f_N$ at $x_0$ gives $\delta>0$ with $|f_N(x)-f_N(x_0)|<\epsilon/3$ whenever $|x-x_0|<\delta$. The <triangle inequality> then gives
$$
|f(x)-f(x_0)|\le |f(x)-f_N(x)|+|f_N(x)-f_N(x_0)|+|f_N(x_0)-f(x_0)|<\epsilon.
$$
Since $x_0$ was arbitrary, $\boxed{f\text{ is continuous on }\mathbb R}$, the <uniform limit theorem>. Continuity of each approximant need not be uniform continuity.
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