Darboux integrability implies Riemann integrability

Here's a direct solution in case you don't want to end up repeating the second half of Riemann-Lebesgue's proof.

You can use the partition $P$ for which $U(f,P)-L(f,P)<\epsilon$ to argue that for all partitions $Q$ with mesh$(Q) = \|Q\|< \delta_P$, the Riemann sum will be somewhere close to $U(f, P)$ and $L(f, P)$.

Assume that $P$ has $N$ points $\{p_1=a, p_2, \cdots, p_N=b\}$, partition $Q$ has $N'$ points $\{q_1=a, \cdots, q_{N'}=b\}$ and $|f| \leq M$ in $[a, b]$ (this should hold for some $M$ or Darboux integral won't be well-defined).

Now take $\delta_P< \min(\|p\|, \frac{\epsilon}{MN})$. Now for each $i \leq N'$, either $[q_i, q_{i+1}] \subset [p_j, p_{j+1}]$ (for some $j\leq N$), or $p_{j-1} \leq q_i \leq p_j \leq q_{i+1} \leq q_{j+1}$. The latter can happen at most $N$ times, so the area under $f$ for such cases can be at most $N \times M \times \delta_P \leq \epsilon$. For the rest, the sum happens to be nicely sandwiched by $U(f,P)$ and $L(f, P)$, proving the Riemann integrability.

Hint

One has anyway $$\lim_{\|P\|\to 0} L(f,P)=\sup_P L(f,P)$$ and $$\lim_{\|P\|\rightarrow 0} U(f,P)=\inf_P U(f,P)$$ If for all $\varepsilon >0$ there exists a partition $P$ such that $U(f,P)-L(f,P)<\varepsilon$, those limits have the same value.

Then $$\lim_{\|P\|\to 0} S(f,P)$$ exists and is equal to that value since $$L(f,P) \le S(f,P) \le U(f,P)$$ for every, however tagged, partition $P$.

The sequential version of the last limit is valid too but one has to suppose $\,\|P_N\| \to 0\,$ of course.