Let $f:[0, t] \to \mathbb R$. For $n \in \mathbb N$, we define $$ \begin{align} V^{(1)}_{n} &:= \sum_{i=1}^{2^n} \left|f\left(\frac{i t}{2^n}\right)-f\left(\frac{(i-1) t}{2^n}\right)\right|, \\ V^{(2)}_{n} &:= \sum_{i=1}^{2^n}\left(f\left(\frac{i t}{2^n}\right)-f\left(\frac{(i-1) t}{2^n}\right)\right)^2. \end{align} $$
Let $V^{(1)} = \lim_n V^{(1)}_{n}$ and $V^{(2)} = \lim_n V^{(2)}_{n}$ be the variation and quadratic variation of $f$ respectively. In proving that the standard Brownian motion has unbounded variation, I come across below result.
Theorem If $f$ is continuous and $V^{(1)} < +\infty$, then $V^{(2)} =0$.
Could you have a check on my below attempt?
Proof Because $[0, t]$ is compact, $f$ is uniformly continuous. Fix $\varepsilon>0$. There is $\delta>0$ such that $|x-y| < \delta$ implies $|f(x) - f(y)| < \varepsilon$. There is $N>0$ such that $\frac{i t}{2^n} - \frac{(i-1) t}{2^n} = \frac{t}{2^n} < \delta$ for all $n\ge N$. Let $$ y_i :=\left|f\left(\frac{i t}{2^N}\right)-f\left(\frac{(i-1) t}{2^N}\right)\right| \quad \forall i=1, \ldots, 2^N. $$
Then $$ \sum_{i=1}^{2^N} y_i^2 \le \bigg (\sum_{i=1}^{2^N} y_i \bigg ) \max \{y_i : i=1, \ldots, 2^N\} \le V^{(1)} \varepsilon. $$
The result then follows by taking the limit $\varepsilon \downarrow 0$. This completes the proof.
As long as you've already shown your expression for $V^{(2)}$ actually gives the quadratic variation of continuous functions, there's only one issue I see to your proof, and it's more a matter of clarity.
When you write $$\sum_{i=1}^{2^N} y_i^2 \le \bigg (\sum_{i=1}^{2^N} y_i \bigg ) \max \{y_i : i=1, \ldots, 2^N\} \le V^{(1)} \varepsilon$$ you have $N$ as a constant depending on $\varepsilon$, so letting $\varepsilon \downarrow 0$ isn't as clean a step as we'd like it to be. Instead, in the setup, you should probably have written $$y_{i,n} :=\left|f\left(\frac{i t}{2^n}\right)-f\left(\frac{(i-1) t}{2^n}\right)\right| \quad \forall i=1, \ldots, 2^n$$ where $n \geq N$ is arbitrary.
Then the statement in question becomes $$\sum_{i=1}^{2^n} y_{i,n}^2 \le \bigg (\sum_{i=1}^{2^n} y_{i,n} \bigg ) \max \{y_{i,n} : i=1, \ldots, 2^n\} \le V^{(1)} \varepsilon$$
The benefit here is that we may first let $n \to \infty$ to show that $$V^{(2)} \leq V^{(1)}\varepsilon,$$ and now letting $\varepsilon \downarrow 0$ to conclude has no possible issue.
To reiterate, you could justify letting $\varepsilon \downarrow 0$ originally without my proposed changes, but you'd need to say more about it. Otherwise, it may appear you're interchanging some limits, which is the source of one of many common errors in analysis.