= Solution
It is enough to treat an increasing function; replacing $f$ by $-f$ handles a decreasing one. Let $\mathcal R$ be the common refinement of dissections $\mathcal D$ and $\mathcal D'$. Refinement raises lower <Darboux sums> and lowers upper ones, so
$$
L_{\mathcal D}(f)\leq L_{\mathcal R}(f)
\leq U_{\mathcal R}(f)\leq U_{\mathcal D'}(f).
$$
For the uniform dissection $\mathcal D_n=\{k/n:0\leq k\leq n\}$, monotonicity gives the telescoping difference
$$
U_{\mathcal D_n}(f)-L_{\mathcal D_n}(f)
=\frac1n\sum_{k=1}^n
\left(f\left(\frac kn\right)-f\left(\frac{k-1}{n}\right)\right)
=\frac{f(1)-f(0)}n.
$$
This can be made smaller than any $\varepsilon>0$, so the <Riemann integrability criterion> proves that $f$ is integrable.
The integral lies between the lower and upper sums, and the displayed sum in the question is the right-endpoint upper sum. Hence the generally valid sharp estimate is
$$
\boxed{
\left|\int_0^1f(x)\,dx
-\frac1n\sum_{k=1}^nf\left(\frac kn\right)\right|
\leq\frac{|f(1)-f(0)|}{n}}. \qquad (1)
$$
The strict inequality printed in the question is false for arbitrary monotone functions: if $f(x)=0$ for $x<1$ and $f(1)=1$, the two sides of (1) are both $1/n$.
For the final claim, write
$$
\Delta_n
=\sum_{k=1}^n\int_{(k-1)/n}^{k/n}
\left(F(x)-F\left(\frac kn\right)\right)dx.
$$
Since $F'$ is continuous on a compact interval, it is uniformly continuous. Uniformly for $0\leq t\leq1$,
$$
F\left(\frac{k-1+t}{n}\right)-F\left(\frac kn\right)
=-\frac{1-t}{n}F'\left(\frac kn\right)+o\left(\frac1n\right).
$$
After the substitution $x=(k-1+t)/n$, summing the uniform errors gives
$$
\Delta_n
=-\frac1{2n^2}\sum_{k=1}^nF'\left(\frac kn\right)
+o\left(\frac1n\right).
$$
The right-hand sum is a <Riemann integral>, so the <fundamental theorem of calculus> yields
$$
\lim_{n\to\infty}n\Delta_n
=-\frac12\int_0^1F'(x)\,dx
=\boxed{\frac{F(0)-F(1)}2}.
$$
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