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.
Let $\mathcal{P}$ ($\mathcal{P}_U$) be the collection of all partitions (uniform partitions) of $[a,b]$.
For a bounded function $f$, we have $L(P,f) \leqslant U(Q,f)$ for lower and upper Darboux sums corresponding to arbitrary partitions $P$ and $Q$. It follows that
$$\sup_{P \in \mathcal{P}}L(P,f) \leqslant \inf_{P \in \mathcal{P}}U(P,f), \quad \sup_{P \in \mathcal{P}_U}L(P,f) \leqslant \inf_{P \in \mathcal{P}_U}U(P,f)$$
Since $\mathcal{P}_U \subset \mathcal{P}$ we have
$$\{U(P,f) \,|\, P \in \mathcal{P}_U\} \subset \{U(P,f) \,|\, P \in \mathcal{P}\}, \quad \{L(P,f) \,|\, P \in \mathcal{P}_U\} \subset \{L(P,f) \,|\, P \in \mathcal{P}\}$$
Hence,
$$\tag{*}\sup_{P \in \mathcal{P}_U}L(P,f) \leqslant \sup_{P \in \mathcal{P}}L(P,f) \leqslant \inf_{P \in \mathcal{P}}U(P,f) \leqslant \inf_{P \in \mathcal{P}_U}U(P,f)$$
Darboux integrability with respect to uniform partitions means that
$$\sup_{P \in \mathcal{P}_U}L(P,f) =\inf_{P \in \mathcal{P}_U}U(P,f),$$
which, in view of (*), implies that $f$ is Darboux integrable with respect to all partitions since
$$0 \leqslant \inf_{P \in \mathcal{P}}U(P,f)- \sup_{P \in \mathcal{P}}L(P,f) \leqslant \inf_{P \in \mathcal{P}_U}U(P,f)- \sup_{P \in \mathcal{P}_U}L(P,f) = 0,$$
and
$$ \inf_{P \in \mathcal{P}}U(P,f)= \sup_{P \in \mathcal{P}}L(P,f) $$
Best Answer
The upper and lower Darboux integrals do not depend on the partitions, they are the infimum and supremum of all upper and lower Darboux sums for all possible partitions of your block. That is, $U(f)=\inf_{P}U(f,P)$ and $L(f)=\sup_{P}L(f,P)$. We know that for any pair of partitions $L(f,P)\leqslant U(f,P')$. If for some partition, $L(f,P)=U(f,P)$, then it follows that $f$ is integrable, for it will follow $L(f)=\sup_P L(f,P)=U(f,P)\geqslant \inf_P U(f,P)=U(f)$. Since we always have $L(f)\leqslant U(f)$, we have $L(f)=U(f)$ and $f$ is Riemann integrable.