You can't base the proof of the statement on an obviously false lemma (at least with an incorrect formulation).
The proof is much simpler than that.
Assume that $u$ is an upper bound for $A$ with the property that, for every $\varepsilon>0$, there exists $a\in A$ with $u-\varepsilon<a$.
By definition, the supremum is the least upper bound. So if $u$ is not the supremum, there is an upper bound $v$ of $A$ with $v<u$.
Now take $\varepsilon=(u-v)/2$. By assumption there exists $a\in A$ with $u-\varepsilon<a$; this becomes
$$
u-\frac{u-v}{2}<a
$$
that is,
$$
\frac{u+v}{2}<a
$$
which is a contradiction, because from $a\le u$ and $a\le v$ we obtain
$$
a\le\frac{u+v}{2}
$$
For the converse, suppose $u=\sup(A)$ and let $\varepsilon>0$. Then $u-\varepsilon<u$, so $u-\varepsilon$ is not an upper bound for $A$, which means there exists $a\in A$ with $u-\varepsilon<a$.
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
Your mistake is thinking you can just subtract inequalities. You can't do that. For example,
$$1>0$$
is true, and $$2>0$$ is also true, but subtracting these two inequalities, I get $$-1>0$$ which is not true.
The problem with subtracting inequalities is that when you subtract equations, you actually multiply one of them by $(-1)$ and then add them. With inequalities, you cannot do that because multiplication by a negative number reverses the inequality.
For an actual proof, a sketch of it would be this:
Naturally, the "near" in this sketch must, in the final proof, be a more rigorous statement. Good luck!