Theorem 6.12(a) Of Baby Rudin. Alternative Proof Of $ \int_a^b \left( f_1 + f_2 \right) d \alpha = \int_a^b f_1 d \alpha + \int_a^b f_2 d \alpha$ 
If $f_1 \in \mathscr{R}(\alpha)$ and $f_2 \in \mathscr{R}(\alpha)$, then
$$ \int_a^b \left( f_1 + f_2 \right) d \alpha = \int_a^b f_1 d \alpha + \int_a^b f_2 d \alpha$$

I want to prove this theorem using definition of integral(upper and lower Riemann-Stieltjes integral). We’ll make use of $ L(P, f_1, \alpha)+ L(P, f_2, \alpha) \leq L(P, f, \alpha) \leq U(P, f, \alpha) \leq U(P, f_1, \alpha) + U(P, f_2, \alpha )$ fact.
My attempt: let $f= f_{1} + f_{2}$.
$\inf U(P,f, \alpha) \leq U(P,f, \alpha) \leq U(P,f_{1}, \alpha) + U(P,f_{2}, \alpha)$. Since $\inf U(P,f, \alpha)$ is an lower bound of $U(P,f_{1}, \alpha) + U(P,f_{2}, \alpha)$, we have $\inf U(P,f, \alpha) \leq \inf \{U(P,f_{1}, \alpha) + U(P,f_{2}, \alpha)\}= \inf U(P,f_{1}, \alpha) + \inf  U(P,f_{2}, \alpha)$. Similarly, $\sup L(P,f_{1},\alpha) + \sup L(P,f_{2},\alpha) \leq \sup L(P,f,\alpha)$.
Thus, $\sup L(P,f_{1},\alpha) + \sup L(P,f_{2},\alpha) \leq \sup L(P,f,\alpha) \leq \inf U(P,f, \alpha) \leq \inf U(P,f_{1}, \alpha) + \inf U(P,f_{2}, \alpha)$. Since, $\sup L(P,f_{1},\alpha) = \inf U(P,f_{1}, \alpha)$ and $\sup L(P,f_{2},\alpha) = \inf U(P,f_{2}, \alpha)$, we have $ \sup L(P,f,\alpha) = \inf U(P,f, \alpha)= \inf U(P,f_{1}, \alpha) + \inf U(P,f_{2}, \alpha)$. Our desired equality $\int_a^b \left( f_1 + f_2 \right) d \alpha = \int_a^b f_1 d \alpha + \int_a^b f_2 d \alpha$.
Is this proof correct?
 A: You snuck in a major step right here that needs an explanation:
$$\inf \{U(P,f_{1}, \alpha) + U(P,f_{2}, \alpha)\}= \inf U(P,f_{1}, \alpha) + \inf  U(P,f_{2}, \alpha)$$
We immediately have
$$\inf \{U(P,f_{1}, \alpha) + U(P,f_{2}, \alpha)\} \geq \inf U(P,f_{1}, \alpha) + \inf  U(P,f_{2}, \alpha)$$
but in the context of the proof, this inequality is pointing in the wrong direction.  What we really need is the other direction.  That is, if we write $\mathcal{P}$ for the set of finite partitions of $[a,b]$, then for clarity, we need to show
$$\inf_{P \in \mathcal{P}} \Big(U(P,f_{1}, \alpha) + U(P,f_{2}, \alpha)\Big) \leq \inf_{P \in \mathcal{P}} U(P,f_{1}, \alpha) + \inf_{P \in \mathcal{P}}  U(P,f_{2}, \alpha).$$

This follows easily from the lemma
$$P_1,P_2 \in \mathcal{P} \implies P_1\cup P_2 \in \mathcal{P} \text { and } U(P_1\cup P_2, f, \alpha) \leq U(P_i,f,\alpha) \text{ for } i=1,2$$
as then for any $P_1,P_2 \in \mathcal{P},$
$$\begin{align*}\inf_{P \in \mathcal{P}} \Big(U(P,f_{1}, \alpha) + U(P,f_{2}, \alpha)\Big) &\leq U(P_1\cup P_2,f_{1}, \alpha) + U(P_1\cup P_2,f_{2}, \alpha) \\ &\leq U(P_1,f_{1}, \alpha) + U(P_2,f_{2}, \alpha)\end{align*}$$
Now by taking the infimum over $P_1,P_2 \in \mathcal{P}$,
$$\begin{align*}\inf_{P \in \mathcal{P}} \Big(U(P,f_{1}, \alpha) + U(P,f_{2}, \alpha)\Big) &\leq \inf_{P_1, P_2\in \mathcal{P}}\Big(U(P_1,f_{1}, \alpha) + U(P_2,f_{2}, \alpha)\Big) \\ &= \inf_{P_1\in \mathcal{P}}U(P_1,f_{1}, \alpha) + \inf_{P_2\in \mathcal{P}}U(P_2,f_{2}, \alpha) \end{align*}$$
[Note that this last equality is the exercise you wanted to use in the comments]

As for the lemma's proof, it should be simple to show $P_1\cup P_2 \in \mathcal{P}$.  For the inequality, you can induct on the size of $P_1\cup P_2$ to assume $|P_1 \cup P_2| = |P_1| + 1$, in which case the proof just amounts to showing that $x_1 < x_2 < x_3$ implies
$$\sup_{x \in [x_1,x_3]}f(x)(\alpha(x_3)-\alpha(x_1)) \geq \sup_{x\in [x_1,x_2]}f(x)(\alpha(x_2)-\alpha(x_1)) + \sup_{x\in [x_2,x_3]}f(x)(\alpha(x_3)-\alpha(x_2))$$
