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 $B$?
You proof is not correct. Since $\varepsilon>0$, you clearly cannot have $M-\varepsilon>M$.
Take $\varepsilon>0$. Then $M+\varepsilon>M$ and therefore $M+\varepsilon$ is not a lower bound of $\{x_n\mid n\in\mathbb N\}$. Therefore, $x_N<M+\varepsilon$, for some $N\in\mathbb N$. Since the sequence is monotonic and decreasing, $n\geqslant N\implies x_n<N+\varepsilon$ too. So$$n\geqslant N\implies x_n\in[M,M+\varepsilon)\implies\lvert x_n-M\rvert<\varepsilon.$$
Best Answer
The idea is fine, but some of the details are wrong. First, by your definition $\sup\{y\in\Bbb R:y\ge x_n\}=+\infty$, which definitely isn’t what you want; from what follows it appears that you want to define $M$ to be the assumed least upper bound of the sequence, i.e., $M=\sup\{x_n:n\in\Bbb Z^+\}$.
Your $\epsilon$ is never actually used. Moreover, if $\epsilon>\frac1M$, choosing $k\in\Bbb Z^+$ so that $x_k>M-\epsilon$ does not ensure that $x_k>M-\frac1M$, since $M-\frac1M>M-\epsilon$. Just observe that the definition of $M$ ensures that there is a $k\in\Bbb Z^+$ such that $x_k>M-\frac1M$. There is no need to rename it $\delta$: that just introduces an unnecessary symbol. Then you can complete the argument roughly as you did, but a bit more efficiently: $x_k<M$, so $\frac1{x_k}>\frac1M$, and therefore
$$x_{k+1}=x_k+\frac1{x_k}>\left(M-\frac1M\right)+\frac1M=M\;,$$
contradicting the definition of $M$.