Given $(f(t)) _{t \ge 0}$, a stochastic process with almost sure continuous paths adapted to the filtration $\mathcal{F}_t = \sigma(W_s : s\le t)$, where $W_s$ is a wiener process.
If we define $$ f_n(t) = \begin{cases} n\int_{\frac{k-1}{n}}^{\frac{k}{n}}f(s)ds, & \frac{k}{n} \le t < \frac{k+1}{n} ~\text{for}~ k = 1, 2, \cdots, n^2-1 \\ 0, & \text{otherwise} \end{cases} $$ why is $(f_n(t))_{t \ge 0}$ a sequence of random step processes? Furthermore, why do we have $$ \lim_{n\to\infty} \int_{0}^N |f(t)-f_n(t)|^2dt = 0, $$ for any fixed $N$ by the continuity path property of $f$?
1) From Jensen, $$\left(\frac{1}{b-a}\int_a^b f\right)^2\leq \frac{1}{b-a}\int_a^b f^2.$$ Take $a=\frac{k-1}{n}$ and $b=\frac{k}{n}$ yields $$\left(\int_{(k-1)/n}^{k/n} f\right)^2=\frac{1}{n^2}\left(n\int_{(k-1)/n}^{k/n}f\right)^2\leq \frac{1}{n}\int_{(k-1)/n}^{k/n}f^2,$$ and thus your result.
2) If $\mathbb E[\int_0^\infty f^2]<\infty $, then $\int_0^\infty f^2<\infty $ a.s. Now, you can prove that if $\int_0^\infty f^2$ converges, then $\lim_{N\to \infty }\int_N^\infty f^2=0$ (which is rather straightforward...)
3) \begin{align*} \int_0^N|f(t)-f_n(t)|^2\,\mathrm d t&=\int_0^N\left|n\int_{(\lfloor tn\rfloor-1)/n }^{\lfloor nt\rfloor/n }f(t)-f(u)\,\mathrm d u\right|^2\,\mathrm d t\\ &\leq \int_0^N\underbrace{\int_{(\lfloor tn\rfloor-1)/n }^{\lfloor nt\rfloor/n }|f(t)-f(u)|^2\,\mathrm d u}_{:=I_n(t)}\,\mathrm d t \end{align*}
where the last inequality follow from 1). Finally, you have that $I_n(t)\to 0$ when $n\to \infty $ and $|I_n(t)|\leq 4\|f\|_{L^\infty (0,N)} $, and thus $(I_n)$ is uniformly bounded. Therefore, Dominated convergence theorem allow you to conclude.