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
At the least, you can conclude that the sublevel sets are pre-compact (I am not sure if they are actually closed). Let us prove this using Arzela-Ascoli.
From Cauchy-Schwarz, you can get equicontinuity (as you have already suggested). In fact you can conclude that for $\gamma \in A_K:=F^{-1}[0,K]$, one has $d(\gamma_s,\gamma_t) \leq \int_s^t |\dot \gamma(s)|ds\leq K|t-s|^{1/2}$.
Thus, we just need to show pointwise relative compactness. Since $X$ is a complete space (hence closed subsets are complete with respect to the same metric), this amounts to showing that for each $t \in [0,1]$, the set $C_{K,t}:=\{\gamma(t):\gamma \in A_K\}$ is totally bounded.
For $R>0$ we define the compact set $L_R:=f^{-1}[0,R]$. Fix a curve $\gamma \in C([0,1],X)$ and let $M>0$. Suppose that $\int_0^1 |\dot \gamma(s)|^2 ds\leq M$, so that $d(\gamma_s,\gamma_t) \leq M|t-s|^{1/2}$ for all $s,t \in [0,1]$. Now fix $t \in [0,1]$ and suppose that dist$(\gamma_t,L_R) =:\delta$; then for all $s \in [0,1]$ we have that dist$(\gamma_s,L_R) \geq \delta - M|t-s|^{1/2}$, and thus $f(\gamma_s) \geq R$ for $s \in [t-(\delta/M)^2,t+(\delta/M)^2]\cap[0,1]$. In particular, $\int_0^1 f(\gamma_s)ds \geq (\delta/M)^2R$. Summarizing this paragraph, we showed that if $\int_0^1 |\dot \gamma(s)|^2ds \leq M$, then for all $R>0$ one has the bound $\int_0^1 f(\gamma_s)ds \geq \sup_{t \in [0,1]}(\text{dist}(\gamma_t,L_R)/M)^2R$.
In particular, if $F(\gamma)\leq K$, then $\int f(\gamma_s)ds \leq K$, therefore setting $M=K$ in the preceding paragraph gives $\sup_{t \in [0,1]} \text{dist}(\gamma_t,L_R) \leq (K/R)^{1/2}K = K^{3/2}/R^{1/2}.$ This bound holds for every $R>0$.
Now we finally prove that $C_{K,t}$ is totally bounded. Fix $K$ and $t$, and let $\epsilon>0$. Choose $R$ large enough so that $K^{3/2}/R^{1/2}<\epsilon/2$. By compactness, there exist $x_1,...,x_N \in L_R$ such that $L_R \subset \bigcup_1^N B(x_i,\epsilon/2)$. Then for any $\gamma \in A_K$, we know dist$(\gamma_t, L_R) < \epsilon/2$, thus $C_{K,t} \subset \bigcup_1^N B(x_i,\epsilon)$. This proves the claim.