There is the following general result:
Let $f: [0,\infty) \to \mathbb{R}$ be a continuous function and $g: [0,\infty) \to \mathbb{R}$ be a function of bounded variation (on compact intervals). Then $$ \langle f, g \rangle_t = 0$$ for all $t \geq 0$, i.e. $$\langle f,g \rangle_t = \lim_{|\Pi| \to 0} \sum_{t_j \in \Pi} (f(t_{j+1})-f(t_j)) (g(t_{j+1}-g(t_j))=0$$ where $\Pi = \{0=t_0< \ldots < t_n = t\}$ denotes a partition of $[0,t]$ with mesh size $|\Pi|$.
Proof: For fixed $t >0$ set
$$\|g\|_{\text{BV}} := \sup_{\Pi} \sum_{t_j \in \Pi} |g(t_{j+1})-g(t_j)|$$
where the supremum is taken over all partitions of the interval $[0,t]$. By assumption, $\|g\|_{\text{BV}}<\infty$. Moreover, because $f$ is uniformly continuous on the interval $[0,t]$, we have
$$\lim_{|\Pi| \to 0} \sup_{|s-r| \leq |\Pi|, s,r \in [0,t]} |f(r)-f(s)| = 0. \tag{1}$$
Consequently,
$$\begin{align*} \left| \sum_{t_j \in \Pi} (f(t_{j+1})-f(t_j)) (g(t_{j+1}-g(t_j)) \right| &\leq \sup_{|r-s| \leq |\Pi|, r,s \in [0,t]} |f(r)-f(s)| \cdot \sum_{t_j \in \Pi} |g(t_{j+1}-g(t_j)| \\ &\leq \|g\|_{\text{BV}} \cdot \sup_{|r-s| \leq |\Pi|, r,s \in [0,t]} |f(r)-f(s)| \\ &\xrightarrow[(1)]{|\Pi| \to 0} 0. \end{align*}$$
By definition, this shows $\langle f, g \rangle_t = 0$.
Back to your original question: Since $X$ has, by assumption, continuous sample paths and $(N_t)_{t \geq 0}$ has sample paths of bounded variation (since the sample paths are non-decreasing), an application of the above statement proves $\langle X, N \rangle = 0$.
I hope the following is an answer in the way you wanted it to be. If there are any mistakes in it, please correct me.
As far as I know, the quadratic variation is only defiened for semimartingales, so we just consider this type of process. As $[X]_0=X_0^2$ we need $|X_0|$ to be deterministic. For simplicity let's focus first on continous processes. Since semimartingales are sums of local martingales and FV-processes, and since the continuous part of the quadratic variation is zero for FV-processes, we can always add a continuous FV-process to any stochastic process with a deterministic quadratic variation without destroying this property. As all contiuous local martingales are time-changed Brownian motions
and since their quadratic variation is the change-of-time, the set contains all time-changed Brownian motions with a deterministic change-of-time (it might sound a bit silly, but this seems to be the answer). Now consider the discontinuous part. As the jumps of the quadratic variation are exactly the squared jumps of the process $\Delta [X]_t=(\Delta X_t)^2$ we need to have deterministic jumps. In summary the set (Hint: This is not unique decomposition.) consists of all processes
$$X=M+A+J$$
(with a deterministic value $|X_0|$), where $M$ is a time-changed Brownian motions with a deterministic change-of-time, $A$ is a continuous FV-process and $J$ is a deterministic pure jump process. $\big($I'm not sure about this: As a non-constant deterministic pure jump process can't be a local martingale, the process $J$ can be included in the (noncontinuous) FV-part, thus $X$ is the sum of a a time-changed Brownian motions with a deterministic change-of-time and a FV-process with deterministic jumps.$\big)$
Another way to look at this is to consider purely discontinuous semimartingales. These are sums of FV-processes and purely discontinuous local martingales, which means that continuous part of the quadratic variation ist zero. As every semimartingale $X$ decomposes uniquely as $$X=X^c+X^d,$$
where $X^c$ is a continuous local martingale with ${X^c_0=0}$ and ${X^d}$ is a purely discontinuous semimartingale, the desired set consists of processes $X$ (with a deterministic value $|X_0|$) which are the sum of a purely discontinuous semimartingale with deterministic jumps and a time-changed Brownian motion with a deterministic change-of-time.
Best Answer
Chapter IV.1 of Revuz and Yor's Continuous Martingales and Brownian Motion, 3rd ed., provides a nice answer. Proposition 1.12 says that for a continuous local martingale $M$,
Thus, if $M$ and $P$ are both continuous local martingales, then choosing $N = M-P$ in your equation yields $[M-P,M-P] = 0$ so that $M$ and $P$ are the same up to a constant. If $M$ and $P$ are instead continuous semimartingales, then they are the same up to a process which is continuous, adapted, and of finite variation (meaning the total variation is finite over every interval $[0,t]$). In other words, their "continuous martingale parts" are the same up to a constant.