= Solution
{c}
Take
$$
\boxed{g(x)=1+|x|,\qquad f(x)=(1+|x|)^2.}
$$
Both <functions> have values in $[1,\infty)$. The <reverse triangle inequality> gives $|g(x)-g(y)|\leq|x-y|$, so $g$ is <Lipschitz continuous> and hence has <uniform continuity>.
For $x_n=n$ and $y_n=n+1/n$, the distances satisfy $|x_n-y_n|=1/n\to0$, whereas
$$
|f(y_n)-f(x_n)|=2\frac{n+1}{n}+\frac1{n^2}>2.
$$
In particular tolerance $\varepsilon=1$ defeats every proposed $\delta$: choose $n$ with $1/n<\delta$. Thus \b[$f$ does not have <uniform continuity> on $\mathbb R$.]
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