If $f$ is K-Lipschitz on a closed interval $I$ and $P_n$ is a partition in $n$ equal parts, then $0\le U(f;P_n)-\int_a^bf\le\frac{K}{n}(b-a)^2$
Proof:
Since $f$ is K-Lipschitz, it satisfies $|f(x)-f(y)|\le K|x-y|\space\forall x,y\in I$ and some $K\in\mathbb{R}>0$. Then let us denote our partition $P_n:=\frac{b-a}{n}$. Hence we want to show $$0\le\sum^n_{k=1}M_k(x_k-x_{k-1})-\lim_{n\to\infty}\sum^n_{k=1}f(x_k)(x_k-x_{k-1})\le\frac{K}{n}(b-a)^2$$ where $M_k=\sup\{f(x):x\in[x_{k-1},x_k]\}$
Here’s where I am stuck, I don’t know how to proceed further. Any help would be appreciated!
You should not write the integral as $$ \lim_{n\to\infty}\sum^n_{k=1}f(x_k)(x_k-x_{k-1}) $$ because $n$ and the partition $x_0, \ldots, x_n$ are fixed.
But you can use that the integral is $\ge$ any lower Darboux sum, in particular $\ge L(f;P_n)$: $$ 0\le U(f;P_n)-\int_a^b f\le U(f;P_n)- L(f;P_n) \\ = \sum^n_{k=1}(M_k-m_k)(x_k-x_{k-1}) $$ with $$ M_k=\sup\{f(x):x\in[x_{k-1},x_k]\} \\ m_k=\inf\{f(x):x\in[x_{k-1},x_k]\} \\ $$ and the Lipschitz condition gives $M_k - m_k \le K(x_k-x_{k-1}) = K \frac{b-a}{n}$.