Confusion on Additive Property of Supremum and Infimum - Theorem I.33 Apostol Calculus I There is a great question and answer to the first part of Theorem I.33 of Apostol Calculus "Additive Property" here. I'm hoping someone can verify my attempt at part (b).
Apostol provides the following theorems

Theorem I.30: *If $x > 0$ and if $y$ is an arbitrary real number, there exists a positive integer $n$ such that $nx > y$
Theorem I.31: If three real numbers a, x and y satisfy the inequalities $$ a \leq x \leq a + y/n$$ *for every integer $n \geq 1$ then $x = a$.
Theorem I.32: Let $h$ be a given positive integer and let $S$ be a set of real numbers.
(a) If $S$ has a supremum, then for some $x$ in $S$ we have$$x > \sup S - h$$
(b) If S has an infimum, then for some $x \text{ in } S$ we have $$x < \inf S + h$$
Theorem I.33 Additive Property: Given nonempty subsets $A$ and $B$ of $\mathbb R$, let $C$ denote the set$$C = \left\{a + b \mid a \in A, b \in B\right\}.$$
(a) If each of $A$ and $B$ has a supremum, then $C$ has a supremum, and $$\sup C  = \sup A + \sup B.$$
(b) If each of $A$ and $B$ has an infimum, then $C$ has an infimum, and $$\inf C  = \inf A + \inf B.$$

Proof of (b)
(1) Assume each of $A$ and $B$ has an infimum. 
(2) If $c \in C$, then $c = a+b$, where $a \in A$ and $b \in B$.
(3) $\forall a \in A, a \ge inf A$, $\forall b \in B, b \ge inf B$ (definition of infimum)
(4) $a + b \ge \inf A + \inf B$, (thm I.25)
(5) $c \ge \inf A +\inf B$ (replacement)
(6) $\inf C \ge \inf A + \inf B.$ ( from (5) and definition of inf)
(7) Let n be any positive integer. $a < \inf A + \frac1n$ and $b < \inf B + \frac1n$ (thm 1.32 with 1/n as h)
(8) $a + b \lt \inf A + \inf B + \frac2n$ (add inequalities, thm I.25)
(9) $a + b - \frac2n \lt \inf A + \inf B$ (thm I.16) 
(10) $\inf C - \frac2n \leq a + b - \frac2n$ (definition of inf and thm I.16)
(11) $\inf C - \frac2n \lt \inf A + \inf B \leq \inf C$ (putting inequalities together)
At this point, the meaning seems right - however we can't use Theorem I.31 in it's current form. So I need to show Theorem I.31 can also be written as $$ a - \frac{y}{n} \leq x \leq a$$
Assume x = a 
done
Assume x < a. 
(1) $(a - x) \in \mathbb{R}^{+}$
(2) $n(a - x) \gt y$ (from thm I.30)
(3) $n(x - a) \lt -y$ (field and order theorems)
This is a contradiction, so $x > a$ cannot hold and $x = a$.
Finally, by Theorem I.31 (reworded) $\inf C = \inf A + \inf B$
Let me know what you think, this feels correct but I wonder if the reworking of Theorem I.31 is necessary?
 A: You're missing a bunch of quantifiers. For example, in your statement (7), what are $a$ and $b$? Is it for all $a$ and $b$? Are they elements of $C$, $A$, $B$, or what?
Of course, I assume you mean that 

"There exist elements $a\in A$ and $b\in B$ such that the inequalities hold."

However you need to make this clear! For example, in inequality (11), what is the quantifier for $n$? Is it a "for all" or a "there exists"?
If we wrote the quantifiers, it should be
(11) For all positive integers $n$, $\inf C-2/n\leq \inf A+\inf B\leq\inf C$.
In particular, if $n$ is any positive integer, then $2n$ is also a positive integer, and we can apply (11) to $2n$ instead and obtain
(11)' For all positive integers $n$, $\inf C-1/n\leq\inf A+\inf B\leq\inf C$.
Which is what you want.
The same type of argument holds for your "modification of Theorem I.31". What is $y$?!

Second, you need to argue why $C$ has an infimum before being able to write $\inf C$ anywhere.

As a last comment, tt seems you want to write a proof in the sense of deductive logic, but this should not be your approach, specially to calculus. Even if you look at Apostol's proofs that is not what he does. Of course, enumerating all your statements could make self-reference easier, however you need to put all quantifiers everywhere, which is time consuming.

Here's a way to write your proof in more standard terms (in a somewhat pedantic way, but this is not a problem: being precise is always better):

Assume that each of $A$ and $B$ has an infimum. In order to prove that $C$ has an infimum, we need to prove that it has a lower bound.
Let $c\in C$ [be arbitrary]. Then $c$ may be written as $c=a+b$, where $a\in A$ and $b\in B$. Then $\inf A\leq a$ and $\inf B\leq b$, so we may add these inequalities to obtain
  $$c=a+b\geq\inf(A)+\inf(B)\tag{1}$$
  Therefore, $\inf(A)+\inf (B)$  is a lower bound of $C$, and thus $C$ has an infimum. Moreover, equation (1) holds for all $c\in C$, and therefore 
  $$\inf(A)+\inf(B)\leq\inf(C)\tag{2}.$$
  Let us prove the converse inequality, and for this we will use Theorem I.31.
Let $n$ be a positive integer. By Theorem I.32(b), there exist $a\in A$ and $b\in B$ such that
  $$a-1/n\leq\inf(A)\leq a\qquad\text{and}\qquad b-1/n\leq\inf(B)\leq b.\tag{3}$$
  Let $c=a+b$, which is an element of $C$ and in particular $\inf C\leq c$. Then adding the two inequalities in (3) yields
  $$\inf(C)-2/n\leq c-2/n\leq\inf(A)+\inf(B).\tag{4}$$
  Adding $2/n$ on the first and last term of (4) gives is $\inf (C)\leq\inf(A)+\inf(B)+2/n$, for any positive integer $n$. Using this with (2) yields
  $$\inf(A)+\inf(B)\leq\inf(C)\leq(\inf(A)+\inf(B))+2/n\quad\text{ for any positive integer }n.$$
  These are precisely the inequalities of Theorem I.31 (using $a=\inf(A)+\inf(B)$, $x=\inf(C)$ and $y=2$), which allows us to conclude that $\inf(A)+\inf(B)=\inf(C)$. Q.E.D

