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.
We can use the following:
Lemma. If $f$ is continuous on $[a,b]$ then for any $\epsilon > 0$ there exists $\delta > 0$ such that for any partition $P$ of $[a,b]$ with $\|P\| < \delta$ and any refinement $R$ of $P$, we have $|S(f,P) - S(f,R)| < \epsilon$.
Applying the lemma, we can show that if $f$ is continuous, then the Cauchy criterion is satisfied. That is, for any $\epsilon >0$ there exists $\delta > 0$ such that if $P$ and $Q$ are any partitions satisfying $\|P\|, \|Q\| < \delta$, then $|S(f,P) - S(f,Q)| < \epsilon.$
To see this, let $R = P \cup Q$ be a common refinement and take $\delta$ as specified in the lemma such that if $\|P\|, \|Q\| < \delta$, we have $|S(f,P) - S(f,R)| < \epsilon/2$ and $|S(f,Q) - S(f,R)| < \epsilon/2$. Whence, it follows that
$$|S(f,P) - S(f,Q)| \leqslant |S(f,P) - S(f,R)| + |S(f,Q) - S(f,R)| < \epsilon/2 + \epsilon/2 = \epsilon.$$
It remains to prove the lemma.
Since $[a,b]$ is compact, $f$ is uniformly continuous and for any $\epsilon > 0$ there exists $\delta >0$ such that if $|x - y| < \delta$, then $|f(x) - f(y)| < \epsilon/(b-a)$. Suppose $\|P\| < \delta$ and $R$ is a refinement of $P$. Any subinterval $[x_{j-1}, x_{j}]$ of $P$ can be decomposed as the union of subintervals of $R$,
$$[x_{j-1},x_{j}] = \bigcup_{k=1}^{n_j}[y_{j,k-1}, y_{j,k}],$$
and
$$\begin{align}\left|S(f,P) - S(f,R)\right| &= \left|\sum_{j=1}^n f(\xi_j)(x_j - x_{j-1}) - \sum_{j=1}^n \sum_{k=1}^{n_j}f(\eta_{j,k})(y_{j,k} - y_{j,k-1})\right| \\ &\leqslant \sum_{j=1}^n \sum_{k=1}^{n_j}|f(\xi_j) - f(\eta_{j,k})|(y_{j,k} - y_{j,k-1}) \\ &\leqslant \sum_{j=1}^n \sum_{k=1}^{n_j}\frac{\epsilon}{b-a}(y_{j,k} - y_{j,k-1}) \\ &= \epsilon\end{align}$$
Best Answer
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.