$\sup (A+B) = \sup A + \sup B$ I realize there are already other duplicates, but none of them are really my question.
In the proof for $\sup (A+B) = \sup A + \sup B$; why is it sufficient to show that $\sup (A +B) + \epsilon \geq \sup A + \sup B$ for all $\epsilon >0$. I already understand the other direction with $\sup (A+B) \leq \sup A + \sup B$
 A: If $$\sup(A+B)<\sup A+\sup B,$$ then there is some $\epsilon>0$ such that $$\sup(A+B)+\epsilon<\sup A+\sup B.$$ Thus, in showing that $$\sup(A+B)+\epsilon\ge\sup A+\sup B$$ for all $\epsilon>0,$ we show that $$\sup(A+B)\ge\sup A+\sup B$$ by contrapositive.

Edit: For a moment, let's forget about the $\epsilon$, and deal with a statement that should be less obscure:

Given real numbers $x,y$. If every real number bigger than $x$ is at least as big as $y$, then $x$ is at least as big as $y$. Symbolically, $$\bigl[\forall z>x\:(z\ge y)\bigr]\Longrightarrow(x\ge y).\tag{$\heartsuit$}$$
Again, we prove this by contrapositive. Suppose $x<y$. Then there is some $x<z<y$. (In particular, letting $z=\frac{x+y}2$ be the midpoint of $x,y$ does the trick.) $\Box$

(The converse of the above holds readily by transitivity of the real order relation.)
Now, $$z>x\quad\Longleftrightarrow\quad z-x>0$$ and $$z\ge y\quad\Longleftrightarrow\quad x+(z-x)\ge y.$$ Letting $\epsilon:=z-x,$ we then have $z>x$ if and only if $\epsilon>0$, and $z\ge y$ if and only if $x+\epsilon\ge y.$ Thus, $(\heartsuit)$ can be rewritten as $$\bigl[\forall\epsilon>0\:(x+\epsilon\ge y)\bigr]\Longrightarrow(x\ge y),\tag{$\star$}$$ or, in words, if $x+\epsilon$ is at least as big as $y$ for all positive real numbers $\epsilon$, then $x$ is at least as big as $y$.
As a further alternative, put into the language of sets and subsets, $(\star)$ can be phrased: "If $(x,\infty)\subseteq[y,\infty),$ then $[x,\infty)\subseteq[y,\infty).$"
A: Edit: I now realize this does not really answer the question, but that was already done several times. Instead, this is how I think you should prove this inequality. No $\epsilon$.
Fix $b\in B$. Then for every $a\in A$
$$
a=a+b-b\leq \sup (A+B)-b \quad\Rightarrow\quad  \sup A\leq \sup (A+B)-b.
$$
So for every $b\in B$
$$
b\leq \sup (A+B)-\sup A\quad\Rightarrow\quad \sup B\leq \sup (A+B)-\sup A.
$$
Note: all this uses is the fact that the sup is the lub. So whenever you exhibit an upper bound, it is not smaller than the sup. Of course, you need to treat the case $A$ or $B$ unbounded above separately: $+\infty=+\infty$. 
A: $A$ and $B$ are sets:
For any $a\in A$ and $b\in B$, we have $a\le\sup(A)$ and $b\le\sup(B)$. Therefore,
$$
a+b\le\sup(A)+\sup(B)
$$
Thus,
$$
\sup(A+B)\le\sup(A)+\sup(B)\tag{1}
$$
$A+B=\{a+b:a\in A\text{ and }b\in B\}$. Thus, for any $a\in A$ and $b\in B$,
$$
a+b\le\sup(A+B)
$$
Taking the $\sup$ over $A$, we get
$$
\sup(A)+b\le\sup(A+B)
$$
then taking the $\sup$ over $B$,
$$
\sup(A)+\sup(B)\le\sup(A+B)\tag{2}
$$
Combining $(1)$ and $(2)$ yields
$$
\sup(A+B)=\sup(A)+\sup(B)
$$

$A$ and $B$ are functions
$$
\sup_{x\in X}(A(x)+B(x))=\sup_{\substack{x\in X\\y\in X\\x=y}}(A(x)+B(y))\le\sup_{\substack{x\in X\\y\in X}}(A(x)+B(y))=\sup_{x\in X}A(x)+\sup_{x\in X}B(x)
$$
The inequality is because the $\sup$ is being taken over a smaller set on the left.
A: If $x$ and $y$ are real numbers, then $x\leq y$ if (and only if) for all $\varepsilon>0$, $x\leq y+\varepsilon$.  
To see this, it is easiest to use contraposition (and there is some repetition of what Cameron Buie posted as I write).  Suppose that $x>y$.  Then for the choice of $\varepsilon =\dfrac{x-y}{2}>0$, we have $x>y+\varepsilon$.
A: For all real numbers $s,t$, the assertion $s+\epsilon \ge t$ for all $\epsilon >0$ is equivalent to the assertion that $s\ge t$ (prove this fact if you are not sure why it is true). The claim you are asking about is a special case of this fact. 
A: You should convince yourself that if $x\geq y$ for all $x\in X$ then $\inf X\geq y$. 
Now, in the spirit of your question, consider
$X=\{\sup(A+B)+\epsilon\;|\;\epsilon>0\}$ and $y=\sup A+\sup B$. It should be immediate that $\inf X=\sup(A+B)$. Therefore, the claim follows. 
