Let $\chi_A$ be the characteristic function of the set $A$.
Then, for any $n\geq 1$,
- $\chi_{C_n}$ is a step function and its integral is equal to $\left(\dfrac{2}{3}\right)^n$ over $[0,1]$.
- $\chi_C\leq \chi_{C_n}$.
- the zero function is a step function lower than $\chi_C$ on $[0,1]$.
So, for any $n\geq 1$, $$0\leq \sup_{u\leq \chi_C} \int_0^1u \leq \inf_{u\geq \chi_C} \int_0^1u \leq \left(\dfrac{2}{3}\right)^n$$
where the $\sup$ and $\inf$ are taken over step functions on $[0,1]$.
Thus the upper integral $U(\chi_C)$ and the lower integral $L(\chi_C)$ of $\chi_C$ are equal and $\chi_C$ is integrable with $$\int_0^1\chi_C=0.$$
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
You can't be sure at this point in your proof, depending on the starting premise, that any other partition $Q$ with mesh $||Q||<\delta$ is necessarily in the set $S$.
What you are trying to prove is usually taken as the definition that $f$ is Riemann integrable.
The function $f:[a,b] \rightarrow \mathbf{R}$ is Riemann integrable with integral value $I$ if for every $\epsilon > 0$ there is a $\delta >0$ such that for any partition $P = (x_0,x_1,\ldots,x_n)$ with $||P|| = \max_{1 \leq i\leq n}(|x_i-x_{i-1}|)< \delta $ and any set of tags $\xi_i \in [x_{i-1},x_i]$ then
$$\left|\sum_{i=1}^{n}f(\xi_i)(x_i-x_{i-1})- I\right|=|S(P,f)-I| < \epsilon.$$
If $f$ is Riemann integrable, then it must be bounded.
So I assume you are trying prove this starting from Darboux's criterion for integrability.
The bounded function $f:[a,b] \rightarrow \mathbb{R}$ is Riemann integrable if the upper and lower Darboux integrals are equal. Or, equivalently, if for any $\epsilon > 0$ there exists a partition $P$ such that $U(P,f)-L(P,f) < \epsilon$
Using your notation:
$$\int_{a}^{b}f=I=U(f) = L(f), $$
where
$$U(f) = \inf_{P} \,U(P,f),\\ L(f)= \sup_{P} \,L(P,f)$$
This implies that for any $\epsilon > 0$ there are partitions $P_1$ and $P_2$ such that $I-\epsilon/2 < L(P_1,f)$ and $U(P_2,f) < I+\epsilon/2.$ Let $P_3 = P_1 \cup P_2$ be a common refinement. Note that a partition $P'$ is a refinement of $P$ if every point in $P$ is also in $P'$. If $P'$ refines $P$ then $||P'|| \leq ||P||$, but the converse is not necessarily true.
Then
$$I-\epsilon/2 < L(P_1,f) \leq L(P_3,f) \leq U(P_3,f) \leq U(P_2,f)< I + \epsilon/2.$$
At this point, you can show that any tagged Riemann sum corresponding to a partition that refines $P_3$ is within $\epsilon$ of $I$, but you still need to show that this is true if the mesh of the partition is sufficiently small regardless of whether or not it refines $P_3.$
Let $D=\sup\{|f(x)−f(y)|:x,y∈[a,b]\}$ denote the maximum oscillation of $f$ and let $δ=ϵ/2mD\,$ where $m$ is the number of points in the partition $P_3$ .
Now let $P$ be any partition with $||P|| < \delta$ . Form the common refinement $Q=P∪P_3$ .
You will see that the upper sums $U(P,f)$ and $U(Q,f)$ differ in at most $m$ sub-intervals and at each the deviation is bounded by $δD$, and
$$|U(P,f)-U(Q,f)| < m\delta D=mD\frac{\epsilon}{2mD}=\epsilon/2$$
It follows that
$$U(P,f)<U(Q,f)+ϵ/2\leq U(P_3,f)+ϵ/2<I+ϵ.$$
By a similar argument, you can show $L(P,f)>I−ϵ$.
Hence for any tagged Riemann sum, $S(P,f)$
$$ I-\epsilon < L(P,f)\leq S(P,f)\leq U(P,f) < I+ϵ,$$
and
$$|I - S(P,f)| < \epsilon.$$