$ A + B $ is defined as $ \{a + b|a \in A, b \in B\} $. To prove the statement in the title, I want to try contradicting $ \sup (A + B) < \sup A + \sup B $.
Here is my thought process: Suppose not $ s = \sup (A + B) $ and $ s < \sup A + \sup B $, then here are the implications of these two assumptions:
\begin{gather} s = \sup (A + B) \implies \forall c \in A + B: c \le s \quad \text{(1)} \end{gather}
\begin{align} s < \sup A + \sup B &\implies ((\forall a \in A, b \in B: a \le u_a \land b \le u_b) \implies s < u_a + u_b) \\ &\implies (s \ge u_a + u_b \implies (\exists a \in A, b \in B: a > u_a \lor b > u_b)) \\ &\implies (s \ge w \implies (\forall x \exists a \in A, b \in B: a > w - x \lor b > x)) && \text{(2)} \\ &(w = u_a + u_b, x = u_b) \end{align}
From (1) and (2) we have:
\begin{gather} \forall c \in A + B: \forall x \exists a \in A, b \in B: a > c - x \lor b > x \end{gather}
It feels like there should be a contradiction here by finding a counterexample for x, but I couldn't find one. What am I doing wrong? Are my assumptions not enough to prove a contradiction?