I think I've proved the statement above, but would like some thoughts if at all possible. I've restated the problem below.
Suppose $f, g$ real and bounded on $[a,b]$. Then show for any partition $\pi$: $U(f+g, \pi) \leq U(f,\pi) + U(g, \pi)$. Where $U$ is the upper sum.
First, I will show for $a \leq x_1 < x_2 \leq b$ that:
$\sup_{x\in(x_1, x_2)} f + g \leq \sup_{x\in(x_1, x_2)}f + \sup_{x\in(x_1, x_2)}g$
Note that since $f, g$ are real and bounded then they have least upper bounds so:
for $x \in (x_1, x_2) \subset [a,b], \ f \leq \sup_{x \in(x_1,x_2)}f$ and $g \leq \sup_{x \in(x_1,x_2)}g$.
Adding these inequalities we have:
$f + g \leq \sup_{x \in(x_1,x_2)}f + \sup_{x \in(x_1,x_2)}g.$
Clearly, $f+g$ is bounded and real and so it has a least upper bound, and by definition we have:
$\sup_{x\in(x_1, x_2)}f + g \leq \sup_{x \in(x_1,x_2)}f + \sup_{x \in(x_1,x_2)}g$.
Now, multiplying by $(x_2 - x_1) > 0$ we get:
$(x_2 - x_1)\sup_{x\in(x_1, x_2)}f + g \leq (x_2 -x_1)\sup_{x \in(x_1,x_2)}f + (x_2 - x_1)\sup_{x \in(x_1,x_2)}g$.
Now summing from $0$ to $n-1$,
$\sum_{i = 0}^{n-1}(x_{i+1} - x_i)\sup_{x\in(x_i, x_{i+1}\ )}f + g \leq \sum_{i = 0}^{n-1}(x_{i+1} -x_i)\sup_{x \in(x_i,x_{i+1} \ )}f + \sum_{i = 0}^{n-1}(x_{i+1} - x_i)\sup_{x \in(x_i,x_{i+1} \ )}g$.
And so, by definition we have:
$U(f+g, \pi) \leq U(f,\pi) + U(g, \pi)$