$S\subseteq T$ means every member of $S$ is a member of $T$.
$a = \inf S$ means $a$ is the largest lower bound of $S$, so it's $\le$ every member of $S$ and $\ge$ all other lower bounds of $S$.
If a number $b$ is $\le$ every member of $T$, and every member of $S$ is a member of $T$, then $b$ is $\le$ every member of $S$. Therefore $\inf T$ is $\le$ every member of $S$. Therefore $\inf T$ is a lower bound of $S$. Therefore $\inf T$ is $\le$ the largest lower bound of $S$. In other words $\inf T\le \inf S$.
A similar argument shows $\sup T\ge\sup S$.
The statement that $\inf S\le \sup S$ is true only if $S\ne\varnothing$. If $S\ne\varnothing$, then there exists $s\in S$. And we must then have $\inf S\le s\le\sup S$.
The fact that if $S\subseteq T$ then $\inf S\ge\inf T$ shows that $\inf\varnothing\ge$ all other "inf"s, so $\inf\varnothing=\infty$. Similarly $\sup\varnothing=-\infty$.
Your question is "how to use this template to show ..." and this is not that hard to answer since $\sup$ and $\inf$ are dual concepts (meaning that the former is the "opposite" of the latter and vice versa). But in order to show this by using the same template one needs to understand the template beforehand (of which I'm not that sure when it comes to you but who am I to judge). So please feel free to ask if you don't understand something. If you want to just have your template for $\inf$ instead of $\sup$, then just skip my explanations in the footnotes.
We want to show that $$\inf(A+B)=\inf A+\inf B.$$
for non-empty and bounded below sets $A,B\subseteq\mathbb{R}$.(1)
First we show that $\inf A+\inf B$ is a lower bound of $A+B$:
$$ \inf A \text{ is a lower bound of } A\text{, so } \alpha \geq \inf A ~~\forall \alpha \in A\\
\inf B \text{ is a lower bound of } B\text{, so } \beta \geq \inf B ~~\forall \beta \in B$$
Thus $\alpha +\beta \geq \inf A+\inf B$ for every $\alpha +\beta \in A+B$, so $\inf A+\inf B$ is a lower bound. $\checkmark$
Second we show that "everything else isn't an lower bound"(2):
For an outline, see (3).
Let $\varepsilon >0$ be given.(4) We have
$$\inf A+\frac{\varepsilon}{2}>\inf A \Rightarrow \exists \alpha\in A\text{ such that } \alpha <\inf A+\frac{\varepsilon}{2}\\
\inf B+\frac{\varepsilon}{2}>\inf B \Rightarrow \exists \beta\in B\text{ such that } \beta <\inf B+\frac{\varepsilon}{2}$$
(5)
So for these $\alpha ,\beta$ we have $$\begin{align*}\alpha +\beta&<\inf A+\frac{\varepsilon}{2}+\inf B+\frac{\varepsilon}{2}\\&=\inf A+\inf B+\varepsilon .\end{align*}$$
So $\inf A+\inf B+\varepsilon$ is not a lower bound. $\checkmark ~~~~\Box$
I hope that I answered as expected and that this won't lead to deeper confusion than you had before.
(1) As they're bounded we don't have to care about the infimum being $-\infty$.
(2) to be precise we show that $\inf A+\inf B$ is the g̲r̲e̲a̲t̲e̲s̲t̲ lower bound.
(3) Short outline of what we're actually doing here: For every $\varepsilon >0$ we show that $\inf A+\inf B+\varepsilon$ is not a lower bound of $A+B$. So there's no lower bound which is greater than $\inf A+\inf B$, so this must be the greatest lower bound.
(4) The identity you stated at this point for the supremum case is obvious because $x-\varepsilon<x$ for arbitrary $x\in\mathbb{R}, \varepsilon>0$. Similarly in the infimum case we have $\inf A+\inf B+\varepsilon>\inf A+\inf B$, but you don't necessarily have to mention that because it's part of the proof strategy as seen in (3).
(5) What's on the left is obvious as before. I don't see how one can deduce from the left to the right side directly but if you know how, then that's okay (probably you have a handy lemma or something). But what's on the right holds anyways because $\inf X$ has the following property: For every $\varepsilon >0$ there exists $x\in X$ such that $x<\inf X+\varepsilon$. (Of course you can use $\varepsilon$/2 for $\varepsilon$.) This holds for arbitrary bounded below $X\subseteq \mathbb{R}$.
Best Answer
This is a little fussy, but we just use the definitions. So, step by step:
if $s=\sup (-A),\ $ then
1). it is an upper bound for $-A$
2). If $x$ is any other upper bound, then $s<x$.
Now you want to show that $-s= \inf A$. So check the conditions:
1). $-s$ is a lower bound for $A$: this follows because $s\geq a$ for all $a\in -A\Rightarrow -s\leq -a$. But $a\in -A\Leftrightarrow -a\in A$, so in fact $-s$ is a lower bound for $A$.
2). Now suppose $y$ is another lower bound for $A$. Then $-y$ is an upper bound for $-A$ (why?) and hence $s<-y$. But then $-s>y$ which finishes the proof.