This is the last step of Rudin's proof for Riesz representation theorem. However I do not understand how he has deducted the (1) and (2) inequalities in the last part of the step. Related question: Riesz Representation Theorem, Rudin: Real and complex analysis
Proof. Clearly, it is enough to prove this for real $f$. Also, it is enough to prove the inequality \begin{equation} \tag{16} \Lambda f \leq \int_X f \,\mathrm{d}\mu \end{equation} for every real $f \in C_c(X)$. For once $(16)$ is established, the linearity of $\Lambda$ shows that $$ -\Lambda f = \Lambda(-f) \leq \int_X (-f) \,\mathrm{d}\mu = - \int_X f \,\mathrm{d}\mu, $$ which, together with $(16)$ shows that equality holds in $(16)$.
Let $K$ be the support of the real $f \in C_c(X)$, let $[a,b]$ be an interval which contains the range of $f$ (note the Corollary to Theorem 2.10), choose $\epsilon > 0$, and choose $y_i$, for $i = 0, 1, \dotsc, n$, so that $y_i - y_{i-1} < \epsilon$ and \begin{equation} \tag{17} y_0 < a < y_1 < \dotsb < y_n = b. \end{equation} Put \begin{equation} \tag{18} E_i = \{ x : y_{i-1} < f(x) \leq y_i \} \cap K \qquad (i = 1, \dotsc, n) \end{equation} Since $f$ is continuous, $f$ is Borel measurable, and the sets $E_i$ are therefore disjoint Borel sets whose union is $K$. There are open sets $V_i \supset E_i$ such that \begin{equation} \tag{19} \mu(V_i) < \mu(E_i) + \frac{\epsilon}{n} \qquad (i = 1, \dotsc, n) \end{equation} and such that $f(x) < y_i + \epsilon$ for all $x \in V_i$. By Theorem 2.13, there are functions $h_i \prec V_i$ such that $\sum h_i = 1$ on $K$. Hence $f = \sum h_i f$, and Step II shows that $$ \mu(K) \leq \Lambda\left( \sum h_i \right) = \sum \Lambda h_i. $$ Since $h_i f \leq (y_i + \epsilon) h_i$, and since $y_i - \epsilon < f(x)$ on $E_i$, we have \begin{align*} \Lambda f &= \sum_{i=1}^n \Lambda(h_i f) \leq \sum_{i=1}^n (y_i + \epsilon) \Lambda h_i \\ &= \sum_{i=1}^n (|a| + y_i + \epsilon) \Lambda h_i - |a| \sum_{i=1}^n \Lambda h_i \\ (1) &\leq \sum_{i=1}^n (|a| + y_i + \epsilon)[ \mu(E_i) + \epsilon/n ] - |a| \mu(K) \\ &= \sum_{i=1}^n (y _i - \epsilon) \mu(E_i) + 2 \epsilon \mu(K) + \frac{\epsilon}{n} \sum_{i=1}^n (|a| + y_i + \epsilon) \\ (2) &\leq \int_X f \,\mathrm{d}\mu + \epsilon[ 2\mu(K) + |a| + b + \epsilon ]. \end{align*}
For $(1)$ note that as shown directly above the displayed formula with $(1)$ it was concluded using Step II that $\sum \Lambda h_i\geq \mu(K)$ holds.
Moreover, as $h_i \prec V_i$ we have $0\leq h_i \leq 1$ with $\mathrm{supp}(h_i)\subseteq V_i$ and so $\Lambda h_i \leq \mu(V_i)$ holds by definition of $\mu$. Then all remains to be done is to use the "defining measure inequality" for the $V_i$.
The main step in $(2)$ is probably using $$y_i-\varepsilon <y_{i-1}<f_{\mid E_i},$$ which shows that the first sum can be estimated by $\int_X f~\mathrm{d}\mu$.
For the second sum, one only estimates $y_i$ by $b$ and calculates the sum.