This is just a matter of using the definitions of supremum and infimum.
Just let $x = \inf{A}$ and $y = \sup{(-A)}$. So since $x = \inf{A}$ then for every $a \in A$ you know that $x \leq a$. But then this implies that $-x \geq -a$ for all $a \in A$, so $-x = - \inf{A}$ is an upper bound for the set $-A$. But since $y = \sup{(-A)}$ is the least upper bound for $-A$, then this means that $y \leq -x$ or that $\sup{(-A)} \leq - \inf{A}$.
You can prove the other inequality.
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
Best Answer
First Part:
You are right that $\frac{1}{\alpha}$ is an upper bound for $\frac{1}{A}$, so $\beta\leq\frac{1}{\alpha}$. However, you also say that $\alpha\leq\frac{1}{\beta}$, but this just implies that $\frac{1}{\alpha}\geq \beta$, which is the same as the previous inequality.
The other inequality that you should use is based on the following. For each $a\in A$ we have $\frac{1}{a}\leq\beta\iff \frac{1}{\beta}\leq a$. Now $\alpha$ is the greatest lower bound for $A$, so we must have $\frac{1}{\beta}\leq\alpha$. This gives $\beta\geq\frac{1}{\alpha}$ which is what we want (combine this with $\beta\leq\frac{1}{\alpha}$ to get $\beta=\frac{1}{\alpha}$).
Second Part: Now if $\alpha=0$, then there is a sequence $(a_n)_{n\geq 1}$ in $A$ with $a_n\searrow 0$. Now $\frac{1}{a_n}\nearrow\infty$ so $\sup(\frac{1}{A})=\infty$.