In the problem above, since f is continuous and thus uniformly continuous on the compact interval $[\frac{\epsilon}{2},1]$, it is possible to directly refer to the general property that all uniformly continuous are Riemann integrable, and then create some partition P of $[\frac{\epsilon}{2},1]$ with all subintervals of length less than $\delta$, which can prove the property.
However, I am interested more in incorporating the hint provided and I want to understand why $\frac{\epsilon}{2}$ was chosen as the left endpoint of interval (rather than say $\epsilon$ or $\frac{\epsilon}{3}$). so I am wondering if it is possible (if so, how) to choose a partition P on $[\frac{\epsilon}{2},1]$ uniquely for the function $f(x)$ shown above that will allow the difference of upper and lower sums to be less than or equal to $\epsilon$ exactly.
Note that $f$ is Riemann integrable on $[a,b]$ iff for any $\epsilon >0$ there exists a partition $P$ such that $$U(P,f)- L(P,f) < \epsilon$$.
For a given $\epsilon>0$, $f$ is Riemann int'ble on $[\epsilon/2 , 1]$(because $f$ is a continuous function on the closed bounded interval $[\epsilon/2 , 1]$). By definition, choose a partition $P_1$ such that $$U(P_1 ,f) -L(P_1 , f) < \epsilon$$
Consider $[0, \epsilon/2]$. Let $P_2 $= {$x_0 = 0, x_1,..., x_n=\epsilon/2$} be a partition.
$$U(P_2 ,f) -L(P_2 , f) = \sum _{i=1}^{n} (M_i - m_i) \triangle {x_i}$$, where $M_i= sup f(x)$ & $m_i = inf (f(x))$ for $x \in [x_{i-1}, x_i]$ for each $i$.
Since $|f(x)|\leq 1$, $M_i - m_i\leq 2$ for each $i=1,2,...,n$.
So,$$U(P_2 ,f) -L(P_2 , f) \leq 2 \sum _{i=1}^{n} \triangle {x_i}=2(x_n -x_0) = 2. \epsilon/2= \epsilon$$
Now, can you choose a parition $P$ so that for a given $\epsilon >0$, $U(P,f)- L(P,f) < \epsilon$??