Okay, so it's uniform equicontinuity. Then there cannot be such an example, since all $f_n$ are uniformly continuous, hence can be continuously extended to $[0,1]$, and the family of extended functions is still uniformly bounded and uniformly equicontinuous. Then by Ascoli's theorem, it contains a uniformly (on $[0,1]$) convergent subsequence.
If we worked with pointwise equicontinuity instead, then
$$f_n(x) = \begin{cases}0 &, x \leqslant 1 - \frac1n\\ n(x-1)+1 &, x \geqslant 1 - \frac1n \end{cases}$$
would give an example.
The family $\{ f_n : n\in\mathbb{Z}^+\}$ is uniformly bounded, and equicontinuous in all $x \in [0,1)$ (in a neighbourhood of each point $x \in [0,1)$, only finitely many $f_n$ are not constant). It converges pointwise to $0$ (and locally uniformly), but not uniformly.
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
Yes, we need. Arzela-Ascoli theorem gives the convergence of some subsequence. The reasoning is the following.