Here is a sketch of a proof that breaks the problem into simpler pieces:
claim 1: If $f$ is bounded with finitely many points of discontinuity on $[a,b]$, then we can write it as $f=f_1+f_2$ where $f_1$ is piecewise constant with finitely many points of discontinuity and $f_2$ is continuous.
claim 2: $f_2\in \mathscr{R}(\alpha)$ by Theorem 6.8.
claim 3: $f_1\in \mathscr{R}(\alpha)$ by the proof of Theorem 6.10; however, the argument is simpler in this case, focusing solely on the discontinuity aspect.
claim 4: $f=f_1+f_2\in \mathscr{R}(\alpha)$ by theorem 6.12
The OP certainly expanded on Rudin's terse style, but one paragraph lacks precision/clarity and needs work:
Because of this, we can find an element $q_1 \in N_1(p) \cap E$, so we can write $q_1 = p_{n_1}$ with $d(p,p_{n_1}) <1$. Now, assume $n_1 <n_2 < \dots <n_{k-1}$ are chosen such that $d(p,p_{n_i}) < 1/i$ for $i = 1,2,3 \dots, k-1$. Then, we can find $q_k \in N_{1/k}(p) \cap E$, meaning that we can write $q_k = p_{n_k}$ for some integer $n_k$ with $d(p,p_{n_k}) < 1/k$. Because the intersection above contains an infinite amount of points, we can certainly find an integer $n_k > n_{k-1}$.
The following is a rework that is more amenable to proof verification:
Define the relation $F_p$ on E by $a \, F_p \, b$ if
$\tag 1 d(b,p) \le (.5) \, d(a,p)$
$\text{and}$
$\tag 2 \text{IF } a = p_n \text{ AND } b = p_m \text{ THEN } m \gt n$
For any $x \in E$ there exist a $y \in E$ with $x \, F_p \, y$. By the axiom of dependent choice, there exist a sequence $q_n$ in $E$ satisfying $q_n \, F_p \, q_{n+1}$. But by (1) it converges to $p$ and by (2), this sequence can also be represented as a subsequence of $p_n$.
I don't think Rudin talks about the axiom of choice in his book.
We encourage the OP to rework his paragraph so that the presentation of variables/notation has a better flow. For example, a sequence is being recursively generated, but the OP's paragraph is kind of fuzzy on $q_2,q_3,\cdots,q_{k-1}$.
The OP might find Some Remarks on Writing Mathematical Proofs / J. M. Lee of interest.
Best Answer
If a non-negative number is smaller than any finite number, it must be zero (you can try to prove it). Rudin actually has a theorem (Theorem 6.6) which says that the necessary and sufficient condition to show that a function is Riemann Integrable is by showing that the difference of the upper and lower integral are smaller than any finite number.