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
$$\begin{align}\left|S(f,P) - S(f,R)\right| &= \left|\sum_{j=1}^n f(\xi_j)(x_j - x_{j-1}) - \sum_{j=1}^n \sum_{k=1}^{n_j}f(\eta_{j,k})(y_{j,k} - y_{j,k-1})\right| \\ &\leqslant \sum_{j=1}^n \sum_{k=1}^{n_j}|f(\xi_j) - f(\eta_{j,k})|(y_{j,k} - y_{j,k-1}) \\ &\leqslant \sum_{j=1}^n \sum_{k=1}^{n_j}\frac{\epsilon}{b-a}(y_{j,k} - y_{j,k-1}) \\ &= \epsilon\end{align}$$
Let $I = \int_a^b f$ and fix $\varepsilon > 0$. From the definition of Riemann integrability, there are partitions $Q_1, Q_2$ of $[a, b]$ such that
$$I-\varepsilon \leqslant \underline{\Sigma}(f, Q_1) \leqslant \overline{\Sigma}(f, Q_2) \leqslant I + \varepsilon.$$
By letting $Q = \{ t_0, t_1, \ldots, t_m \}$ (where $a = t_0 < t_1 < \cdots < t_m = b$) be the common refinement of $Q_1, Q_2$, we get
$$I-\varepsilon \leqslant \underline{\Sigma}(f, Q) \leqslant \overline{\Sigma}(f, Q) \leqslant I + \varepsilon.$$
Since $f$ is bounded, there is $M$ such that $|f(x)| \leqslant M$ for all $x \in [a, b]$. Let $p$ be so large that
(i) $(m-1) \cdot \frac{b-a}{p} \cdot M \leqslant \varepsilon$,
(ii) $\frac{b-a}{p} \leqslant \displaystyle \min_{1 \leqslant i \leqslant m} (t_i - t_{i-1})$.
Now fix $n \geqslant p$. We will show that
$$I - 3\varepsilon \leqslant \underline{\Sigma}(f, P_n) \leqslant \overline{\Sigma}(f, P_n) \leqslant I + 3\varepsilon,$$
which will complete the proof.
Let $R$ be the common refinement of $Q$ and $P_n$. We're going to show that $R$ has just a little greater inferior sum than $P_n$. By (ii) each interval in $P_n$ contains at most one point $t_i$, unless it contains two, in which case they are it's endpoints. So for each $k = 1, 2, \ldots, m-1$: either $t_k \in P_n$, or adding $t_k$ to $P_n$ divides some interval $I_k$ in $P_n$ into two proper subintervals $I_k = I_k^- \cup I_k^+$ (and $I_k \neq I_l$ for $k \neq l$). Let $S$ be the set of those $k$ for which the second case holds. Then it's easy to see that
$$\begin{align*}
\underline{\Sigma}(f, R) - \underline{\Sigma}(f, P_n) & = \sum_{k \in S} \left[ |I_k^-| \cdot \inf_{x \in I_k^-} f(x) + |I_k^+| \cdot \inf_{x \in I_k^+} f(x) - |I_k| \cdot \inf_{x \in I_k} f(x) \right] \\[1ex]
& \leqslant \sum_{k \in S} \left[ |I_k^-| \cdot M + |I_k^+| \cdot M + |I_k| \cdot M \right] = 2M \sum_{k \in S} |I_k|.
\end{align*} $$
But of course there are at most $m-1$ points in $S$ and $|I_k| = \frac{b-a}{n} \leqslant \frac{b-a}{p}$. Hence, by (i)
$$\underline{\Sigma}(f, R) - \underline{\Sigma}(f, P_n) \leqslant 2M \cdot (m-1) \cdot \frac{b-a}{p} \leqslant 2 \varepsilon.$$
Also since $R$ is a refinement of $Q$, $\underline{\Sigma}(f, R) \geqslant \underline{\Sigma}(f, Q)$. So we get
$$\underline{\Sigma}(f, P_n) \geqslant \underline{\Sigma}(f, R) - 2\varepsilon \geqslant \underline{\Sigma}(f, Q) - 2 \varepsilon \geqslant I - 3\varepsilon.$$
We analogously show that $\overline{\Sigma}(f, P_n) \leqslant I + 3 \varepsilon$, which concludes the proof.
Best Answer
I'll answer your question about how we can assume $\delta_{n} \geq \delta_{n+1}$. (You say that you understand why we want it to be true.)
Let's fix $n$ for the moment. Let's say we already found $\delta_1,\ldots, \delta_n$. According to the hypothesis, we can find $\delta_{n+1}^{\ast} > 0$ such that $$ |S(f,P) - S(f,Q)| < \frac{1}{n+1} $$ for any tagged partitions $P$ and $Q$ with $\|P\| < \delta_{n+1}^{\ast}$ and $\|Q\| < \delta_{n+1}^{\ast}$.
I used the notation $\delta_{n+1}^{\ast}$ because it is our first try for $\delta_{n+1}$. It could be that $\delta_{n+1}^{\ast}$ is larger than $\delta_{n}$.
To fix that, for our second attempt we define $$ \delta_{n+1} = \min(\delta_n,\delta_{n+1}^{\ast}). $$ Let's check that $\delta_{n+1}$ does what we want.
Clearly $\delta_{n} \geq \delta_{n+1}$. So that's taken care of.
We also want $\delta_{n+1}$ to have the property that $$ |S(f,P) - S(f,Q)| < \frac{1}{n+1} $$ for any tagged partitions $P$ and $Q$ with $\|P\| < \delta_{n+1}$ and $\|Q\| < \delta_{n+1}$. Let's check it.
From the definition of $\delta_{n+1}$, it's clear that $\delta_{n+1}^{\ast} \geq \delta_{n+1}$. So for any tagged partitions $P$ and $Q$ with $\|P\| < \delta_{n+1}$ and $\|Q\| < \delta_{n+1}$, we have $\|P\| < \delta_{n+1}^{\ast}$ and $\|Q\| < \delta_{n+1}^{\ast}$, and therefore (by the original property of $\delta_{n+1}^{\ast}$) we have $$ |S(f,P) - S(f,Q)| < \frac{1}{n+1}. $$