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
$$|I - S(P,f)| < \epsilon.$$
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
$$
|S(f,P) - S(f,Q)| < \frac{1}{n+1}.
$$
Best Answer
The theorem you refer to is called Cauchy's condition for Riemann integrability and is more on the lines of Cauchy's general principle of convergence.
The theorem related to difference between upper and lower Darboux sums is called Riemann's condition for Riemann integrability and it was explicitly mentioned by Riemann while developing a theory of his integral.
These two theorems are different in the sense that one deals with Riemann sums (the Cauchy's condition) and the other deals with Darboux sums.
Regarding your question about publishing I have no idea as I haven't published any paper. But I think if one explicitly mentions the terms used (including names of theorems) this would remove any ambiguity and should not be objected.