Understanding the proof of the Arzelà–Ascoli theorem from wikipedia This is the statement and the proof from wikipedia.
I have highlighted the part of the proof that i have questions with.
Let I = [a, b] ⊂ R be a closed and bounded interval. If F is an infinite set of functions  f  : I → R which is uniformly bounded and equicontinuous, then there is a sequence fn of elements of F such that fn converges uniformly on I.
Proof:
Fix an enumeration ${x_i}_{i}$ of rational numbers in I. Since F is uniformly bounded, the set of points ${f(x_1)}, f∈F$ is bounded, and hence by the Bolzano-Weierstrass theorem, there is a sequence ${f_{n1}}$ of distinct functions in F such that ${f_{n1}(x_1)}$ converges. Repeating the same argument for the sequence of points ${f_{n1}(x_2)}$, there is a subsequence ${f_{n2}}$ of ${f_{n1}}$ such that ${f_{n2}(x_2)}$ converges.
By induction this process can be continued forever, and so there is a chain of subsequences
$f_{n1} \supseteq f_{n2} \supseteq ...$
such that, for each k = 1, 2, 3, ..., the subsequence ${f_{nk}}$ converges at $x_1, ..., x_k$. Now form the diagonal subsequence ${f}$ whose $m$th term $f_m$ is the $m$th term in the $m$th subsequence ${f_{nm}}$. By construction, $f_m$ converges at every rational point of I.
Therefore, given any ε > 0 and rational xk in I, there is an integer N = N(ε, xk) such that
$
|f_n(x_k) - f_m(x_k)| < \tfrac{\varepsilon}{3}, \qquad n, m \ge N.
$
Since the family F is equicontinuous, for this fixed ε and for every x in I, there is an open interval Ux containing x such that
$|f(s)-f(t)| < \tfrac{\varepsilon}{3}$
for all $f ∈ F$ and all $s, t $ in I such that $s, t ∈ U_x$.
The collection of intervals $U_x, x ∈ I$, forms an open cover of I. Since I  is compact, this covering admits a finite subcover $U_1, ..., U_J$. There exists an integer K such that each open interval $U_j, 1 ≤ j ≤ J$, contains a rational $x_k$ with $1 ≤ k ≤ K$. Finally, for any $t ∈ I$, there are j and k so that $t$ and $x_k$ belong to the same interval $U_j$. For this choice of k,
\begin{align} \left |f_n(t)-f_m(t) \right| &\le \left|f_n(t) - f_n(x_k) \right| + |f_n(x_k) - f_m(x_k)| + |f_m(x_k) - f_m(t)| \\ &< \tfrac{\varepsilon}{3} + \tfrac{\varepsilon}{3} + \tfrac{\varepsilon}{3} \end{align}
for all $n, m > N = max{N(ε, x1), ..., N(ε, xK)}$. Consequently, the sequence ${f_n}$ is uniformly Cauchy, and therefore converges to a continuous function, as claimed. This completes the proof.
Why is the above statement true ? I cannot derive it from the fact that $f_n$ is equicontinous.
 A: The first sentence, namely 

(1) Therefore, given any $\def\e{\varepsilon}\e > 0$ and rational $x_k \in I$, there is an integer $N = N(\e, x_k)$ such that $|f_n(x_k)−f_m(x_k)|<\frac \e3$, for all $n,m\ge N$. 

follows from the fact that the sequence $\bigl(f_n(x_k)\bigr)_n$ for fixed $k$ is convergent (the part before the highlighted sentence), hence Cauchy. Now for the second sentence, where equicontinuity comes into play:

(2) Since the family $F$ is equicontinuous, for this fixed $\e$ and for every $x\in I$, there is an open interval $U_x$ containing $x$ such that $|f(s)−f(t)|< \frac \e 3$, all $s,t \in U_x \cap I$.

That follows from the very definition of equicontinuity, namely: Given $\e > 0$, choose $\delta > 0$ such that
$$ |f(s) - f(x)| < \frac \e6, \quad |s-x| < \delta,\quad f \in F $$
Now let $U_x = (x -\delta, x+\delta)$, then $U_x$ contains $x$ and by the above, we have for $f \in F$ and $s,t \in U_x$ that 
$$ |f(s) - f(t)| \le |f(s) - f(x)| + |f(x) - f(t)| < \frac \e 3 $$
