For example, take the case where we want to show $\sup B= 1 - \inf A$. We have already shown that $1-\inf A$ is an upper bound, so $1 - \inf A \geq \sup B$.
For the other way, pick $\epsilon > 0$. We want to show that there is some $n$ and some $a \in A$ with $\frac{n}{n+1} - a < 1 - \inf A- \epsilon$.
This is simple : note that we can find $N$ so that $1-\frac{N}{N+1} < \frac\epsilon 2$(take any $N > \frac 2 \epsilon$), and then we can find $a \in A$ such that $a - \inf A < \frac{\epsilon}{2}$ (by the definition of infimum).
Add these up and rearrange to get $\frac{N}{N+1}- a> 1 - \inf A - \epsilon$.
In other words, if $A$ AND $n$ are involved, then split the given $\epsilon$ into smaller $\epsilon$-numerator fractions, obtain separate equations for $A$ and $n$ and then combine them.
I leave you to figure out how the second one can be done. Remember, obtain separate equations for $A$ and $n$ and then combine them.
This is the sort of situation where generality helps.
Result : For any two subsets $X$ and $Y$ of the real line, define $X+Y = \{x + y : x \in X, y \in Y\}$. If $X,Y$ are bounded, then so is $X+Y$. Furthermore, we also have the following formulas : $\inf X + \inf Y = \inf(X+Y)$, and $\sup(X+Y) = \sup X + \sup Y$.
Proof : I will do it for the supremum, you figure out the infimum : it is exactly the same.
For any $z \in X+Y$, we know $z = x+y$ for some $x\in X,y \in Y$. Therefore $z \leq \sup X + \sup Y$. It follows that $\sup X + \sup Y$ is an upper bound for $X+Y$, so $\sup X+Y \leq \sup X + \sup Y$.
For the other way, fix $\epsilon > 0$. Let $ x',y' $ be such that $\sup X - x' < \frac \epsilon 2$ and $\sup Y - y' < \frac \epsilon 2$. Add and conclude that $(\sup X + \sup Y) - (x'+y') < \epsilon$. Therefore, $\sup X+Y = \sup X + \sup Y$ .
Result : If $A$ is bounded, then $-A$ is bounded, with $\sup (-A) = - \inf (A)$ and $\inf (-A) = -\sup A$.
Prove this yourself.
Now, just note for your question that $B = S + (-A)$, where $A$ is some bounded set and $S = \{\frac{n}{n+1} : n \in \mathbb N\} \cup \{0\}$. Can you use the general result to find the infimum and supremum of $B$?
To show that $5$ is the upper bound, show that each successive term is decreasing (i.e. $\frac{5}{n+1}< \frac5n$). So no term exceeds $5$, so $5$ is the least upper bound.
To show that $0$ is the greatest lower bound, first you know that $0$ is a lower bound since you're dividing positive numbers so $5/n$ can never be negative. Then show that for any $\epsilon>0$, you can find some $n$ such that $5/n<\epsilon$. So $0$ is the greatest lower bound.
Best Answer
You have come up with good candidates for the supremum and the infimum. The strategy of proof is this. Let the candidate for the supremum be $s$. You first show that $s$ is an upper bound for the set. Then you pick an arbitrary $\epsilon \gt 0$ and show that there exists an element $a \in A$ such that $s \lt a$. For example,
$ \frac{n - k}{n + k} \lt 1$ and hence $1$ is an upper bound for $A$. Let $\epsilon $ be arbitrary. Fix $k$. You can pick $n \in \Bbb N$ such that $\frac{2k}{n + k} \lt \epsilon $. Then, $1 - \epsilon \lt 1 - \frac{2k}{n + k} = \frac{n - k}{n + k} $ and you are done.
For the infimum, you need to show that your candidate $d$ is a lower bound for the set and then if you pick a point which is at an arbitrarily small distance to the right of $d$ on the line, there is an element $a' \in A$ which is less than that point.
First, you can note that $ \frac{|n - k|}{|n + k|} = \frac{|n - k|}{n + k} \lt 1 $ and hence by definition, $-1$ is a lower bound for the set $A$. Now as before let $\epsilon \gt 0$ be arbitrary. We need to show there exists an element $a' \in A$ such that $a' \lt -1 + \epsilon $. Again pick $ \frac{2k}{n + k} \lt \epsilon $ as before which can be done since $n$ can be made arbitrarily large and you will be done.