Let $g:[a,b]\to[c,d]$ be continuous and $f:[c,d]\to\mathbb{R}$ be Riemann-integrable.
Why is it not possible to show by means of Darboux-sums that $(f\circ g)$ is Riemann-integrable?
The sketch of a fake-proof could be as follows:
As $f$ is Riemann-integrable there exists a partition $P$ of $[c,d]$ such that $U(f,P)-L(f,P)<\epsilon$. Let be $\delta>0$ the width of the smallest subintervall $[t_{i-1},t_i]$ of $[c,d]$ which is obtained by the partition $P$.
Let be $P':=\{t_0,t_1,\cdots, t_n\}$ a partition of $[a,b]$ such that for all $i\in \{1, 2,\cdots,n\}$ and $x,y \in [t_{i-1},t_i]$ it follows (due to uniform continuity of $g$): $$|g(x)-g(y)|\leq |\sup\{g(x)\mid x\in[t_{i-1},t_i]- \inf\{g(x)\mid x\in[t_{i-1},t_i]\}|<\delta.$$
Now let's define $$M_i:=\sup\{g(x)\mid x\in[t_{i-1},t_i]\}\\m_i:=\inf\{g(x)\mid x\in[t_{i-1},t_i]\}.$$
So I can simply construct a partition $P'':=\{d_0,d_1,\cdots d_m\}$ of $g([a,b])$ by using the finite many $M_i$'s and $m_i$'s. For example, a $d_j$ can be either $d_j=\sup\{g(x)\mid x\in[t_{i-1},t_i]\}$ or $d_j=\inf\{g(x)\mid x\in[t_{i-1},t_i]\}$. It follows that for all $j\in\{1,2,\cdots,m\}$ we have $|d_{j-1}-d_j|\leq\sup\{g(x)\mid x\in[t_{i-1},t_i]\}-\inf\{g(x)\mid x\in[t_{i-1},t_i]\}<\delta$.
So the width of every subinterval which is obtained by the partition $P'$ of $[c,d]$ is smaller than $\delta$. Hence, $U(f\circ g, P')-L(f\circ g, P')<U(f,P)-L(f,P)<\epsilon$ and the composition would be Riemann-integrable.
Where are my mistakes?