Let $t<u$ and $\pi = \{0=t_0 < t_1 < ... < t_{m(\pi)}=t\}$ of $[0,t]$, where mesh$(\pi)=\max_i(t_{i+1}-t_i)$.
Given that $[X,Y]_u$ and $[X,Y]_t$ satisfy the limit \begin{align} [X,Y]_t = \lim_{\text{mesh}(\pi) \to 0} \sum_i (X_{t_{i+1}}-X_{t_{i}})(Y_{t_{i+1}}-Y_{t_{i}}) \qquad \text{in probability.} \end{align}
To show that, \begin{align} (*) [X,Y]_u - [X,Y]_t = \lim_{\text{mesh}(\pi) \to 0} \sum_i (X_{s_{i+1}}-X_{s_{i}})(Y_{s_{i+1}}-Y_{s_{i}}) \qquad \text{in probability,} \end{align} taken over partitions $\{s_i\}$ of $[t,u]$ as the mesh tends to $0$.
I started with, \begin{align} &[X,Y]_u - [X,Y]_t \\ &= \lim_{\text{mesh}(\pi) \to 0} \sum_i (X_{u_{i+1}}-X_{u_{i}})(Y_{u_{i+1}}-Y_{u_{i}}) - (X_{t_{i+1}}-X_{t_{i}})(Y_{t_{i+1}}-Y_{t_{i}}). \end{align}
Now, for me it is unclear how to choose the partition of $[t,u]$ such that equality $(*)$ holds.
Fix $t \leq u$ and $\epsilon>0$. By the very definition of $[X,Y]_t$ and $[X,Y]_u$, we can choose $\delta>0$ such that
$$\left| [X,Y]_t - \sum_{t_i \in \Pi} (X_{t_{i+1}}-X_{t_i})(Y_{t_{i+1}}-Y_{t_i}) \right| \leq \epsilon \tag{1} $$
and
$$\left| [X,Y]_u - \sum_{t_i \in \Pi'} (X_{t_{i+1}}-X_{t_i})(Y_{t_{i+1}}-Y_{t_i}) \right| \leq \epsilon \tag{2}$$
for any partition $\Pi$ and $\Pi'$ of $[0,t]$ and $[0,u]$, respectively, with mesh size $\leq \delta$. Now let $\tilde{\Pi}$ be an arbitrary partition of $[t,u]$ with mesh size $\leq \delta$. Define a partition of $[0,t]$ by
$$\Pi' := \left\{ 0 = t_0 \leq t_1 := \frac{t}{n} \leq t_2 := 2 \frac{t}{n} \leq \ldots \leq t_n := t \right\}$$
where $n$ is chosen sufficiently large such that $\text{mesh}(\Pi') \leq \delta$. Obviously, $\Pi := \Pi' \cup \tilde{\Pi}$ is a partition of $[0,u]$ with mesh size $\leq \delta$. Since
$$\left| [X,Y]_u-[X,Y]_t - \sum_{t_i \in \tilde{\Pi}} (X_{t_{i+1}}-X_{t_i})(Y_{t_{i+1}}-Y_{t_i}) \right| = \left| [X,Y]_u-[X,Y]_t - \left( \sum_{t_i \in \Pi} - \sum_{t_i \in \Pi'} (X_{t_{i+1}}-X_{t_i})(Y_{t_{i+1}}-Y_{t_i}) \right) \right|$$
it follows from $(1)$, $(2)$ and the triangle inequality that
$$\left| [X,Y]_u-[X,Y]_t - \sum_{t_i \in \tilde{\Pi}} (X_{t_{i+1}}-X_{t_i})(Y_{t_{i+1}}-Y_{t_i}) \right| \leq 2 \epsilon.$$
Since $\epsilon>0$ is arbitrary, this finishes the proof.