You are almost correct. Let $\epsilon >0$ and $P$ be a partition of $[a, b]$ such that $(M+1)||P||< \epsilon$ (see definition of $M$ below). Then
$$S(P) - s(P) = \sum_i^n (M_i - m_i) \Delta x_i\ .$$
By definition of $M_i$ and $m_i$, there exists $a_i$ and $b_i$ in $[x_i, x_{i+1}]$ such that
$$|f(a_i) - f(b_i)| > M_i - m_i - 1/n$$
Then
$$S(P) - s(P) < \sum_{i=1}^n (|f(a_i) - f(b_i)| + 1/n) \Delta x_i \leq ||P||\big(\sum_{i=1}^n (|f(a_i) - f(b_i)|\big) +||P||$$.
Let all these $a_i$ and $b_i$ be a subset of some partition of $[a, b]$, as $f$ is of bounded variation, $\sum_i (|f(a_i) - f(b_i)|<M$ for some $M$ (which depends only on $f$). Thus
$$S(P) - s(P) < (M+1)||P|| < \epsilon\ .$$
Thus $f$ is Riemann integrable. Indeed, there is a result showing that every function of bounded variation can be written as the difference of two increasing functions. As increasing functions are Riemann integrable, so are functions of bounded variations.
See this Bounded variation, difference of two increasing functions for the last fact mentioned.
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
$$|I - S(P,f)| < \epsilon.$$
Best Answer
By "integrable" you appear to mean "Riemann-integrable", i.e. you're using partitions of an interval and upper and lower sums. In that sense of integrability, not all bounded functions are integrable. For example, let $$ f(x) = \begin{cases} 1 & \text{if $x$ is rational,} \\ 0 & \text{otherwise.} \end{cases} $$ This is bounded and not Riemann-integrable. And it does not have bounded variation on any interval. This shows that having bounded variation is a far stronger condition than merely being bounded. And that merely being bounded is nowhere near enough for your purpose.