Define $I(X)=\int_{0}^{T}X(s)dB(s)$ and we see that $E(I(X)^2)=E\int_{0}^{T}X^2(s)ds.$ If $E(I(X)^2)=0$, then $\int_{0}^{T}X^2(s)ds=0$ and it should be $X(s)=0$ a.s. for any $0\leq s\leq T$. But there is a counterexample $X(s)=1_{[0,a]}(s)+1_{(a,\tau]}(s)\frac{1}{T-s},$ where $\tau=inf\{t:t\geq a, \int_{a}^{t}\frac{1}{T-s}dB(s)=-B(a)\}$, which was taught in class. It has $I(X)=0$ but $X\neq 0$.
The homework states that $\int_{0}^{t}X(s)dB(s)=0$ for any $t\in [0,T]$ implies $X=0$. Is it safe to use the square integral approach similarly mentioned above? Or is there any conditions I'm missing?
Sorry, I misunderstood: One can show $\tau \le T$. The condition $\mathbb{E}[I(X)^2] = 0$ does not imply $X(s) = 0$ for all $s$, as the counterexample shows. The reason is that the identity $\mathbb{E}[I(X)^2] = \mathbb{E}[\int_0^T X(s)^2ds]$ you referenced is only true provided both sides are finite. In the counterexample you showed, we have $\mathbb{E}[\int_0^T X(s)^2ds] = \infty$, so it does not need to equal $\mathbb{E}[I(X)^2]$.
To show that the correct condition is $\int_0^t X(s)dB(s) = 0$ for all $t \in [0,T]$ implies $X(s) = 0$ for all $s \in [0,T]$, the square integral approach won't work for the same reason as it didn't for the earlier counterexample. Instead, I would suggest using that a constant martingale has a quadratic variation of $0$.