Let's look at the upper bounds of $A = \{(x,y) \in\mathbb{R}^2 | 1 \le x \le 2\}$. You already saw that $(3,0)$ is an upper bound, because $3>2$, but so are each $(3,-1)>(3,-2)>(3,-3)>\ldots$, so we need to find smaller upper bounds. The same holds if we replace $3$ above with some $a>2$.
Trying $a=2$ doesn't work, since for any $b\in\mathbb{R}$, we have that $(2,b)<(2,b+1)\in A$, so $(2,b)$ is not a canditate for the supremum. Obviously the there are no upper bounds with $a<2$.
So we are left with the set $\{(a,b)\mid a>2,b\in\mathbb{R}\}$ of upper bounds of $A$. The problem here is that we cannot find a smallest element in this set, as you can always choose $2+\frac{a-2}{2}$, which is smaller than $a$ (even if you could find this $a$, then we could still make $b$ approach $-\infty$, meaning that there is still no smallest element).
Analogously, $A$ doesn't have an infimum.
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 $B$?
Best Answer
This is mostly right, but I think you should be a little more explicit about where you get the contradiction from. In particular, how do you know that there is an element $r \in A$ such that $M<r$? This doesn't come from the fact that $A$ is bounded, it's because $M < \sqrt{2}$.