Because if you have a uniformly convergent subsequence and the space is closed your subsequence will converge in the space, then you have a that for any sequence you can extract a convergent subsequence in the space, which is one of the characterizations of a compact space.
If the space is equicontinuous any sequence inside it is equicontinuous.
Same as above. Or: $$\{f_n\} \subseteq \{f\in\mathcal{F}\}$$
so
$$\sup_{f_n}\||f_n||_{\infty} \leq \sup_{f\in\mathcal{F}}||f||_{\infty}<\infty $$
For each $x$ he has some subsequence $f_{n_i}(x_i)$ that converges pointwise, taking the intersection of all the $n_i$ and renaming as $n$ he gets a sequence that converges for every $x_i$
Let's first try to develop an idea on how to start the proof. Let's assume for simplicity that $\alpha (x)=x$. We are given that $f:[a,b]\to \mathbb R$ is discontinuous at finitely many points in $[a,b]$.We want to show that $f$ is RS$(\alpha)$ integrable. Now, given any $\epsilon\gt 0$, we want to be able to choose a partition $P_\epsilon$ of $[a,b]$ so that $U(P_\epsilon, f,\alpha)-L(P_\epsilon, f, \alpha)\lt \epsilon$. So let $x_1\lt x_2\lt x_3$ be points in $[a,b]$ where $f$ is discontinuous. Let's "capture" these $x_i$'s in very small intervals. So let's choose $u_i\lt x_i\lt v_i$ such that $v_i-u_i\lt\epsilon/3$ and of course $u_i,v_i$ belong to $[a,b]$. (Note that sum of lengths of these intervals is less than $\epsilon/3+\epsilon/3+\epsilon/3=\epsilon$). Note that the set $K:=[a,b]-\cup_{i=1}^3(u_i,v_i)$ is compact. $P:=\{a\le u_1\lt v_1 \lt u_2\lt v_2\lt u_3\lt v_3\le b\}$ is a partition of $[a,b]$. To take advantage of uniform continuity over $K$, we "refine" $P$ to $P_\epsilon$ by adding points outside $[u_i,v_i]$'s.
Based on the idea developed above, let's proceed with the proof. Let $E=\{x_1,x_2,...,x_m\}\subset [a,b]$ be the points of discontinuity of $f$ in $[a,b]$and assume that $\alpha$ is monotonically increasing. Let $\epsilon\gt 0$ be arbitrary.
Claim: Since $\alpha $ is continuous at $x_i$, we can find $u_i\lt v_i$ such that $\alpha(v_i)-\alpha(u_i)\lt \frac {\epsilon}m. \tag 1$
Proof: Let $\{l_n\}$ (such that $[a,b]\ni l_n\lt x_i$ for all $n\in \mathbb N$) be a sequence converging to $x_i$; and let $\{r_n\}$ be sequence in $[a,b]$ such that $r_n\gt x_i$ for all $n$ and that $r_n$ converges to $x_i$. It follows by continuity of $\alpha$ at $x_i$ that $\alpha (l_n)\to \alpha(x_i)$ and $\alpha (r_n)\to \alpha (x_i)$. It follows that $\alpha (r_n)-\alpha (l_n)\to 0.$ So for some large enough $N$, we must have $\alpha (r_N)-\alpha (l_N)\lt \frac\epsilon m$. Set $v_i=r_N$ and $u_i=l_N$. $∎$
Now, note that $K:=[a,b]-\cup_{i=1}^m(u_i,v_i)$ is compact. $f$ must be uniformly continuous on $K$. There exists a $\delta\gt 0$ such that for $x\in K \land y\in K, |x-y|\lt \delta\implies |f(x)-f(y)|\lt \epsilon$.
So we refine $P$ by adding points $y_j$'s outside $[u_i,v_i]$ such that
None of the $y_j$'s lies in $[u_i, v_i]$ for any $i$.
$y_j-y_{j-1}\lt \delta\iff [y_{j-1},y_j]\cap [u_i,v_i]=\emptyset$
If $j$ is the largest index such that if $y_j\lt u_i$ (where $i$ is the smallest index satisfying this) then $u_i-y_j\lt \delta$; and similarly if $j$ is the smallest index such that $y_{j-1}\gt v_i$ then $y_{j-1}-v_i\lt \delta$.
Let $P':=$ union of these $y_j$'s with $u_i$'s and $v_i$’s. Let $1\le j\le n$ for some appropriate $n$.
Now $U(P',f,\alpha)-L(P',f,\alpha)=\sum_{i=1}^m (M_i-m_i)(\alpha (v_i)-\alpha(u_i))+\sum_{i=1}^n(M_i-m_i)(\alpha(y_i)-\alpha(y_{i-1})$
The first term on RHS is $\le 2M \sum_{i=1}^m (\alpha (v_i)-\alpha (u_i))\lt 2M \epsilon$ (by $(1)$).
The second term on RHS is $\lt \epsilon \sum_{i=1}^n(\alpha(y_i)-\alpha(y_{i-1})\leq \epsilon (\alpha (b)-\alpha (a))$.
So finally, we get: $U(P,f,\alpha)-L(P,f,\alpha)\lt \epsilon ((\alpha (b)-\alpha (a))+2M)$, whence the conclusion follows.
Best Answer
Now that you have updated with the filled in lines, it seems the confusion is mixing up the conditions for continuity and convergence. Uniform continuity has nothing to do with taking $n≥N$; that would be convergence. And the line ending with an asterisk has nothing to do with the convergence. What you can do is take $N$ different $\delta$s corresponding to the uniform continuity of each of the first $N$ functions, then take their minimum. As you said, it is possible because there are only finitely many.
Notice that for convergence, you are talking about functions getting closer together at each point (uniformly in this case) as $n$ gets larger, whereas for continuity you are talking about each function taking close values at nearby points. Equicontinuity says that the you can choose your definition of "nearby" the same for all functions in the family simultaneously. The step you are asking about is basically saying that a finite family of uniformly continuous functions is equicontinuous.