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
$$\begin{align}\left|S(f,P) - S(f,R)\right| &= \left|\sum_{j=1}^n f(\xi_j)(x_j - x_{j-1}) - \sum_{j=1}^n \sum_{k=1}^{n_j}f(\eta_{j,k})(y_{j,k} - y_{j,k-1})\right| \\ &\leqslant \sum_{j=1}^n \sum_{k=1}^{n_j}|f(\xi_j) - f(\eta_{j,k})|(y_{j,k} - y_{j,k-1}) \\ &\leqslant \sum_{j=1}^n \sum_{k=1}^{n_j}\frac{\epsilon}{b-a}(y_{j,k} - y_{j,k-1}) \\ &= \epsilon\end{align}$$
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.
Best Answer
Fix $n$ large. Then for $1/n\le x <y\le 1,$ the mean value theorem shows
$$\tag 1 |f(y)-f(x)|\le n^2|y-x|.$$
Now let $m\in \mathbb N$ be greater than $n^3.$ Set
$$P_n=\{0\} \cup\{1/n+k\frac{(1-1/n)}{m},k=0,1,\dots ,m\}.$$
Using $(1),$ we get
$$U(f,P_n)-L(f,P_n)\le \frac{2}{n} +\sum_{k=1}^{m} \left (n^2\cdot \frac{1-1/n}{m}\right)\cdot \frac{1-1/n}{m}$$ $$ \le \frac{2}{n} + m\frac{n^2}{m^2} < \frac{2}{n} + \frac{1}{n} = \frac{3}{n}.$$