I have an intricate issue with the diagonalization argument used in the proof of Arzela-Ascoli theorem. It goes as follows:
So assume that $\scr F$ has these three properties [closed, bounded, equicontinuous] and let $(f_n)$ be a sequence in $\scr F$. We will construct a convergent subsequence. By Corollary 8.6.8, there is a sequence $(x_i)$ such that for each $r > 0$, there is an integer $N$ such that $\{x_1, \dots, x_N\}$ forms and $r$-net for $K$.
We claim that there is a subsequence of $(f_n)$, call it $(f_{n_k})$, such that $$\lim_{k \to \infty} f_{n_k}(x_i) = L_i\quad\text{exists for all }i \ge 1.$$ To prove this, let $\Lambda_0$ denote the set of positive integers. Since $\big(f_n(x_1)\big)$ is a bounded sequence, the Bolzano-Weierstrass Theorem (2.7.2) provides a convergence subsequence. That is, there is an infinite subset $\Lambda_1\subset \Lambda_0$ such that $$\lim_{n \in \Lambda_1} f_n(x_1) = L_1\quad\text{exists.}$$ Next, $\big(f_n(x_2)\big)_{n\in\Lambda_1}$ is a bounded sequence, so there is an infinite subset $\Lambda_2 \subset \Lambda_1$ such that $$\lim_{n \in \Lambda_2} f_n(x_2) = L_2\quad\text{exists.}$$ Continuing in this way, we obtain a decreasing sequence $\Lambda_0 \supset\Lambda_1 \supset\Lambda_2 \supset\cdots$ of infinite sets such that $$\lim_{n \in \Lambda_i} f_n(x_i) = L_i\quad\text{converges for each }i \ge 1.$$
We now use a diagonalization method, similar to the proof that $\Bbb R$ is uncountable, Theorem 2.9.8. Let $n_k$ be the $k$th entery of $\Lambda_k$; and let $\Lambda = \{n_k : k \in \Bbb N\}$. Then for each $i$, there are at most $i-1$ entries of $\Lambda$ that are not in $\Lambda_i$. Thus $$\lim_{k\to\infty} f_{n_k}(x_i) = \lim_{n \in \Lambda_i} f_n(x_i) = L_i\quad\text{for all}\quad i \ge 1,$$ proving the claim.
For simplicity of notation, we use $g_k$ for $f_{n_k}$ in the remainder of the proof. Now fix $\varepsilon > 0$. By uniform equicontinuity, there is $r > 0$ such that $$\|f(x) - f(y)\| < \dfrac{\varepsilon}3\quad\text{for all}\quad f\in \mathscr F\text{ and }\|x-y\|< r.$$ Choose $N$ such that $\{x_1, \dots,x_N\}$ is an $r$-net for $K$. Since the $g_k$ converge at each of the $N$ points, there is some $M$ such that $$\|g_k(x_i) - g_l(x_i)\| < \dfrac{\varepsilon}3\quad\text{for all}\quad k,l \ge M\text{ and }l \le i \le N.$$
Let $k,l \ge M$ and pick $x \in K$. Since $\{x_1, \dots, x_N\}$ is an $r$-net for $K$, there is some $i \le N$ such that $\|x - x_i\| < r$. We need an $\varepsilon/3$-argument to finish the proof: $$\begin{align}\|g_k(x) - g_l(x)\|&\le \|g_k(x) - g_k(x_i)\| + \|g_k(x_i) - g_l(x_i)\| + \|g_l(x_i) - g_l(x)\|\\ &\le\dfrac{\varepsilon}3+\dfrac{\varepsilon}3+\dfrac{\varepsilon}3 = \varepsilon\end{align}$$ Thus $g_k$ is uniformly Cauchy, and so converges uniformly by Theorem 8.2.2. The limit $g$ belongs to $\scr F$ because $\scr F$ is closed. Finally, since every sequence in $\scr F$ has a convergent subsequence, it follows that $\scr F$ is compact.
Initially my question was:
As far as I understand, the diagonalization argument has $r\rightarrow0$ and $N\rightarrow\infty$ which will map every point in the set $K$ to some sequence. But real numbers on any set like $[a,b]$, $a<b$ are uncountable and by taking $N\rightarrow\infty$ we assume they are countable? The proof maps every point in set K to some sequence, correct? The proof presumes we can count these sequences since we will take $k$-th element from $k$-th sequence but the way these sequences are constructed and mapped to in the diagonalization argument signals we can't. Am I missing something?
After a discussion with Paul Sinclair below, I figured out I missed the meaning of Corollary 8.6.8 somehow and $r\rightarrow0$ and $N\rightarrow\infty$ is not even considered in the proof. The sequence will be the same but for every $r$-net we might take a head of the sequence of different length $N$ every time.
Not so much "missing something" as "seeing things that are not there".
You have read far too much into what is just an off-hand comment in the proof. Note that the author says "a diagonalization method", not "a diagonalization argument". The difference is that author is just building a sequence. They are not making an argument about countability or uncountability at all. The comment was just to give another example where a sequence was built by following a diagonal. The purposes of these two sequences are entirely different. And the conclusion obtained from this diagonalization is the proof of the opening claim: the existence of a subsequence of functions $f_{n_k}$ such that for all $i, \lim_{k \to \infty} f_{n_k}(x_i)$ converges. the line "proving the claim" at the top of the second page signals the end of anything related to diagonalization.
As already noted, the diagonalization bit is already finished at this point, but the proof also doesn't have $r \to 0$ and $N \to \infty$ in it. Instead, it fixes an arbitrary $\varepsilon > 0$, and uses that to fix a value of $r$ per equicontinuity. Then it uses that fixed $r$ to fix a value of $N$. It is true that a smaller $\varepsilon$ will require a smaller $r$, and that in turn will require a larger $N$. But that is immaterial to the argument being made. That argument only requires that for any given $\varepsilon$, you can find a single corresponding value of $r$ and a single corresponding value of $N$.
No. Just, no. There is no such claim. There is not even a suggestion of any such thing in the proof. I can only presume you came up with the idea based on the misunderstanding of what they meant by the reference to Cantor's diagonalization argument.
The rationals form an $r$-net for $\Bbb R$ for every value of $r > 0$. But no one suggests that means $\Bbb R$ is mapped to $\Bbb Q$. It obviously is not.
The rest of your post just builds on these false ideas.
As an aside - not a problem with your overall argument, just an easily fixed incorrect detail - compactness does not guarantee that a set of real numbers is uncountable. All finite sets are compact. Many countable sets, such as $\{0, \frac 1n \mid n \in \Bbb N^+\}$, are also compact. What assures $[a,b]$ is uncountable is that it contains $(a,b)$, and all non-empty open sets of real numbers are uncountable.