Prove $\sup(f+g) \le \sup f + \sup g$

Consider $x\in D$. Then $f(x)\le \sup f$ and $g(x)\le \sup g$, hence $(f+g)(x)=f(x)+g(x)\le \sup f+\sup g$. Therefore, $\sup f + \sup g$ is some upper bound, but the least upper bound may be smaller.

Your example for strictness is fine.

Another example for < from (http://people.reed.edu/~davidp/212.2013/handouts/sup_example.pdf)

The pdf doesn't expatiate on this, but $\sup f(x) + g(x) = 1 $ at both $x = -1, 1$.
pdf simply chagrins about $x = 1$. But at $x = -1$, $f(-1) + g(1) = 1 + 0 = 1$.