Prove $\sup(f+g) \le \sup f + \sup g$ 
Suppose $D$ is a nonempty bounded subset of reals. Let $f:D \to \mathbb R$ and $g:D \to \mathbb R$. Define $(f+g)(x)=f(x)+g(x)$. Prove $\sup(f+g)(D) \le \sup f(D) + \sup g(D)$ (also prove that $\sup (f+g)$ exists).

I understand why this is the case, just not how to prove it. Left side is pretty much $\sup  (f(x)+g(x))$ and right side is $\sup (f(x) + g(y))$ for $x,\,y \in D$. Basically $f+g$ has to use the same variable and $f(D)+g(D)$ use different ones. But I don't know how to go about proving this.

The second part of the question is to find a specific example where strict inequality holds. 

Let $D=\{a,b\}$ and $f: a \to 1,\, b\to 0, \,g: a \to 0,\, b\to 1$. $\sup f(D) = 1,\, \sup g(D) = 1,\, \sup f(D) + \sup g(D) = 2.$ $\sup (f+g)(D) = 1$ (if we choose a, $f+g = 1+0,\, b,\, f+g=0+1$).
 A: 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 $x = -1$, $f(-1) + g(1) = 1 + 0 = 1$. 

A: 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.
A: Another example for < from Abbott
The two sides are usually not equal because the functions f and g could
easily take on their larger values in different places of each subinterval. For
example, consider f(x) = x and g(x) = 1 − x on the interval [0, 1]. Then
$\sup\{f(x) : x ∈ [x_{k−1}, x_k]\} = 1$,
$\sup\{g(x) : x ∈ [x_{k−1}, x_k]\} = 1$, 
but $\sup\{\color{seagreen}{f(x) + g(x) = 1} : x ∈ [x_{k−1}, x_k]\} = \sup\{\color{seagreen}{1}\} =  1 \neq 2$
A: To show that $\sup(f+g)\leq \sup(f) +\sup(g)$, suppose that $\sup(f)=A$ and $\sup(g)=B$, where $A$ and $B$ are the highest point of function $f$ and $g$ respectively.
If we add $f$ and $g$, then only two posibility happen:
1: If $A$ and $B$ are in same x, then $f(x)=A$ and $g(x)=B$. Then, we have that $(f+g)(t)=A+B$. Hence, $$\sup(f+g)=\sup(f)+\sup(g).$$
2: If $A$ and $B$ are not in same $x$, then $f(p)=A$ and $\ g(q)=B$. Hence, $$\sup(f+g)\leq A+B$$
From 1 and 2, we have that $$\sup(f+g)\leq\sup(f)+\sup(g).$$
