Let $\Pi_n$ be a sequence of partitions with $|\Pi_n| \to 0$. Then the $p$-variation of a continuous function $g$ along the partitions $\Pi_n$ is defined as
$$V_T^p(g) = \lim_{n \to \infty} V_T^p(g, \Pi_n) = \lim_{n \to \infty} \sum_{t_i \in \Pi_n} |g(t_i \wedge T) - g(t_{i-1} \wedge T)|^p$$
Since $g$ is continuous,
$$\forall \varepsilon > 0 \ \exists\ \delta(\varepsilon): |t_i - t_{i-1}| < \delta(\varepsilon) \implies |g(t_i \wedge T) - g(t_{i-1} \wedge T)| < \varepsilon$$
Now since $|\Pi_n| \to 0$, for every $\varepsilon > 0 \ \exists \ \Pi_n : |\Pi_n| < \delta(\epsilon)$ which implies
$$\sum_{t_i \in \Pi_n} |g(t_i \wedge T) - g(t_{i-1} \wedge T)|^p < \sum_{t_i \in \Pi_n} \varepsilon^p \le M\varepsilon^p$$
Where $M$ is the (finite) number of points in the partition $\Pi_n$ (the number of points is finite because by definition $\Pi_n \cap [0,t]$ is finite for every $t$)
Therefore, since the sum is less that $M\varepsilon^p$ for every positive $\varepsilon$, we conclude that it is $0$.
Except that it is not, as not all continuous function have bounded variation!
So what is the error?
This would be true if we also add the hypothesis that $g$ has finite $r$ variations, for $r < p$. But how does the hypothesis enter the argument, and what is wrong with what I've written above?