As an exercise in my book, there are a few questions that should guide us through the fact that if $f$ is continuous and $g$ is Riemann integrable on a compact interval, then $f(g)$ is also Riemann integrable.
- Show that if $ f$ is 1-Lipschitz then $f(g)$ is Riemann integrable.
- Show that every continuous function can be written as a uniform limit of 1-Lipschitz (!) functions.
- Deduce that if $f$ is continuous then $f(g)$ is Riemann integrable.
Did the author maybe mean "uniform limit of Lipschitz functions"? Because if I am not mistaken, $\sqrt x$ cannot be represented as uniform limit of 1-Lipschitz functions on any compact neighborhood of $0$.
As commenters said, it is impossible to approximate a general continuous function by $1$-Lipschitz functions, since the $1$-Lipschitz property passes to limits (even pointwise ones).
However, on a compact metric space $X$, it is possible to approximate a continuous function $f$ uniformly by Lipschitz functions $f_n$ with Lipschitz constant depending on $n$. This is enough for the above proof.
Such an approximation can be defined in one line: $$f_n(x) = \sup_{y\in X} (f(y)-nd(x,y)).$$ This is evidently $n$-Lipschitz, and $f_n\ge f$. When $n>2\delta^{-1}\sup_X|f|$, the supremum is in effect taken over $d(x,y)\le \delta$, which yields $f_n\to f$ uniformly.