Solution 1:

  1. As you mentioned in a comment, you should choose $C_i \in \Big[ \inf f_1 \restriction R_i, \sup f_1 \restriction R_i \Big]$. This way you get $\big\|f_1 -f_2\big\|_\infty < \delta$. An adequate choice of $\delta > 0$ (that is $\delta |R|^{\frac{1}{p}} < \varepsilon$) would then give you $\big\| f_1 - f_2\big\|_p \leq \varepsilon$.
  2. $$ \big\| f_1 - f_2\big\|_p = \left(\int_R |f_1 - f_2|^p \right)^{\frac{1}{p}} \\ \leq \left( \int_R \big\| f_1 - f_2\big\|_\infty^p \right)^{\frac{1}{p}} \\ \left(|R|\cdot \big\| f_1 - f_2\big\|_p \right)^{\frac{1}{p}} \\ |R|^{\frac{1}{p}} \cdot \big\| f_1 - f_2\big\|_\infty.$$
  3. A metrizable space is separable if and only if it is second-countable. This means that the euclidean topology on $\mathbb{R}^n$ has a countable basis. This countable basis is explicitely described in the first paragraph of the proof and is put to use in the second paragraph.