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 $x_i$
You can find the Arzelà–Ascoli in utmost generality (definitely more than you need) in Engelking's "General Topology", theorems 3.4.20 (p. 163) and 8.2.10 (p. 443). In your case, these theorems reduce to:
$F \subset C([0, \infty))$ is relatively compact in the topology of uniform convergence on compact subsets of $[0, \infty)$ if and only if $F$ is equicontinuous at every point $x \in [0, \infty)$ and $\{f(x) \mid f \in F\} \subseteq \mathbb R$ is bounded for every $x \in [0, \infty)$.
Minor nitpick: it is not clear if $N$ in your question is supposed to be natural; even if it were, you may easily replace it with real positive numbers, because every real number $x$ sits between its integer part $[x]$ and $[x]+1$, which are natural numbers. (In fact, $N$ plays the role of the interval $[0,N]$, which can be readily replaced with arbitrary compacts of $[0, \infty)$.)
Pick $x \in [0, \infty)$ arbitrary. In relation (1) take $N=x$. From the $\varepsilon - \delta$ definition of the concept of limit, taking $\varepsilon = 1$ there exists $\delta_1 > 0$ such that if $\delta < \delta_1$ then $\sup _{f \in F} V^x (f, \delta) \le 1$. Equivalently, $V^x (f, \delta) \le 1$ for all $f \in F$ and $\delta < \delta_1$. Explicitly,
$$\sup \{ |f(s) -f(t)| : 0 \le s, t \le x, \ |s-t| < \delta \} \le 1 \quad \forall f \in F \ .$$
In particular, taking $\delta = \frac {\delta_1} 2 < \delta_1$ in the above, we get that $|f(s) -f(t)| \le 1 $ for all pairs $0 \le s, t \le x$ with $|s-t| < \frac {\delta_1} 2$ and for all $f \in F$.
The interval $[0,x]$ can be covered with $n(x, \delta_1) := \left[ \frac {2x} {\delta_1} \right] + 1$ subintervals of length $\frac {\delta_1} 2$, whence (using the triangle inequality multiple times)
$$|f(x) - f(0)| \le \\
\le \left| f(x) - f \left( x - \frac {\delta_1} 2 \right) + f \left( x - \frac {\delta_1} 2 \right) - f \left( x - 2 \frac {\delta_1} 2 \right) + \dots + f \left( x - n(x, \delta_1) \frac {\delta_1} 2 \right) - f(0) \right| \le \\
\le \left| f(x) - f \left( x - \frac {\delta_1} 2 \right) \right| + \left| f \left( x - \frac {\delta_1} 2 \right) - f \left( x - 2 \frac {\delta_1} 2 \right) \right| + \dots + \left| f \left( x - n(x, \delta_1) \frac {\delta_1} 2 \right) - f(0) \right| \le \\
\le 1 + 1 + \dots + 1 = n(x, \delta_1) \ .$$
(Warning: my count above might be off by $\pm 1$ because of the endpoints, I'm never good at counting, but this doesn't change the end result: we have been able to find an upper bound for $|f(x) - f(0)|$ which is independent of $f \in F$.)
Finally, if $B = \{ |f(0)| \mid f \in F\}$ then
$$|f(x)| = |f(x) - f(0) + f(0)| \le |f(x) - f(0)| + |f(0)| \le n(x, \delta_1) + |f(0)| \in n(x, \delta_1) + B$$
and the right-hand side is obviously bounded, being the translate by the constant (with respect to $f \in F$) $n(x, \delta_1)$ of the bounded subset $B$.
(Notice that since $\{f(0) \mid f \in F\}$ was bounded, so will be $\{|f(0)| \mid f \in F\}$, trivially.)
Since $x$ was arbitrary, all the work above proves the pointwise boundedness of $F$.
The (uniform, but this is not needed) equicontinuity of $F$, on the other hand, comes practically for free, being encoded in (1), as you remark yourself in the question.
Since you have pointwise boundedness and equicontinuity, you have relative compactness in the topology of uniform convergence on compact subsets.
Best Answer
Recall that for all $f \in C(X)$, $f$ is uniformly continuous because $X$ is compact.
Given $F \subset C(X)$ a finite set, we show that it is equicontinuous. Fix $\varepsilon > 0$. We need to find som $\delta > 0$ such that some condition is satisfied.
For all $f \in F$, by uniform continuity of $f$, there exist $\delta_f$ such that etcetera. Your required $\delta$ is $\delta = \min_{f \in F} \delta_f$.
In particular, using the same argument, you can see that a finite union of equicontinuous sets is equicontinuous.