$f$ is a is continuous and has bounded variations, then for every sequence $Π^n$ of subdivisions of $[0,T]$ such that $|Π^{n}|:=sup |t_i - t_{i-1}| → 0$ then how to prove that $$\lim \sup_n \sum^n_{k=1} |f_{t_k}-f_{t_{k-1}} |^2 = 0$$
P.s I don not know how to fix the situation of the square of the variation.
$\def\vps{\varepsilon}$ Suppose $f$ is continuous and is of bounded variation on the interval $[0,T]$ and let $M$ be the supremum of all the sums $$ \sum^n_{k=0} |f(t_k)-f(t_{k-1})|, $$ where $0=t_0<t_1<\cdots<t_q=T$. You may assume $M>0$ without loss of generality.
Since $f$ is continuous on a bounded closed interval, it is uniformly continuous on $[0,T]$.
Let $\vps>0$. By uniform continuity, there is $\delta>0$ such that $|f(x)-f(y)|<\vps/M$ whenever $|x-y|<\delta$.
By assumption, there is $N\in\mathbb N$ such that $\max_k|t_k - t_{k-1}|<\delta$ for any subdivision $Π^n: 0=t_0<t_1<\cdots<t_q=T$ whenever $n \ge N$. Now for any of the subdivisions $Π^n$ with $n\ge N$ we have $$ \sum^q_{k=0} |f(t_k)-f(x_{t-1})|^2 \le \sum^q_{k=0} |f(t_k)-f(x_{t-1})|\cdot \vps /M = M\cdot \vps/M = \vps. $$ This proves the required limit.