If $t\mapsto X_t$ continuous, why $X_t^{(n)}\to X_t$ in $L^2$? Let $(\Omega ,\mathcal F,\mathcal P)$ a probability space. Let $(X_t)_t$ a bounded stochastic process s.t. for all $\omega $, the function $t\mapsto X_t(\omega )$ is continuous. Let $$X_t^{(n)}(\omega )=X_0(\omega )\boldsymbol 1_{\{0\}}(t) +\sum_{k=0}^{2^n-1}X_{\frac{kT}{2^n}}(\omega )\boldsymbol 1_{\left(\frac{kT}{2^n},\frac{(k+1)T}{2^n}\right]}(t).$$ 
Why $X_t^{(n)}\to X_t$ in $L^2(\Omega \otimes [0,T])$ ?
Attempts 
Fix $t\geq 0$. Let $t_n\in \mathbb N$ s.t. $\frac{t_nT}{2^n}<t\leq \frac{(t_n+1)T}{2^n}.$ Then $$X_t^{(n)}(\omega )=X_{t_n}(\omega ),$$
for all $n$, and thus, by continuity, $$\lim_{n\to \infty }X_{t}^{(n)}(\omega )=\lim_{n\to \infty }X_{t_n}(\omega )=X_t(\omega ).$$ 
So we have $X_t^{(n)}(\omega )\to X_t(\omega )$ for all $\omega $. Since $X_t$ and $X_n$ are bounded, by bounded convergence theorem, the claim follow. 
Question : Is this correct ? And if yes, why did they considered 
$$X_t^{(n)}(\omega )=X_0(\omega )\boldsymbol 1_{\{0\}}(t) +\sum_{k=0}^{2^n-1}X_{\frac{kT}{2^n}}(\omega )\boldsymbol 1_{\left(\frac{kT}{2^n},\frac{(k+1)T}{2^n}\right]}(t)$$
and not   $$X_t^{(n)}(\omega )=X_0(\omega )\boldsymbol 1_{\{0\}}(t) +\sum_{k=0}^{n-1}X_{\frac{kT}{n}}(\omega )\boldsymbol 1_{\left(\frac{kT}{n},\frac{(k+1)T}{n}\right]}(t) \ \ ?$$

As you have proven, $X_t^{(n)}(\omega)\to X_t(\omega)$ for all $\omega$ and $t\in[0,T]$ as $n\to\infty$. Using Fubini's theorem and bounded convergence implies
\begin{align*}
\lim_{n\to\infty}\int_\Omega\int_0^T \left|X_t^{(n)}-X_t\right|^2\,dt\,dP&=\lim_{n\to\infty}\int_0^T \int_\Omega\left|X_t^{(n)}-X_t\right|^2\,dP\,dt\\&=\int_0^T\lim_{n\to\infty}\int_\Omega\left|X_t^{(n)}-X_t\right|^2\,dP\,dt\\&=\int_0^T 0\,dt=0.
\end{align*}
Hence $X_t^{(n)}\to X_t$ in $L^2(\Omega\times[0,T],P\otimes\lambda)$ where $\lambda$ is the Lebesgue measure. 
Finally, you can also consider
$$X_t^{(n)}(\omega )=X_0(\omega )\boldsymbol 1_{\{0\}}(t) +\sum_{k=0}^{n-1}X_{\frac{kT}{n}}(\omega )\boldsymbol 1_{\left(\frac{kT}{n},\frac{(k+1)T}{n}\right]}(t) $$
which has the same property. 
