So for the supremum of a set, we say:
$S=sup(A) \Leftarrow \Rightarrow S - \epsilon < x$ for some $x \in S$ and $\epsilon > 0$.
I don't understand this for specific examples where the supremum is not in set A, such as:
$A=\{1-1/n:n \in \mathbb{N} \}$
Here the supremum would be $S = 1$, but if $\epsilon = 0.00000... 0001$, it would be greater than 0, but $S - \epsilon \not < x$ for some $x \in S$ (They would be the same).
Your definition of $sup(A)$ is not quite right.
$S = sup(A) \iff \forall \epsilon > 0 ~\exists x ~\in ~A ~\text{such that}~ (S - \epsilon) < x.$
This means, per the definition of $sup(A)$,
first any (small positive) $\epsilon$ is chosen,
then a satisfying $x ~\in A$ must be found.
From the definition of $sup$ that I have given you, you can see that the example that you offered is not in fact a counter example.
That is, in your example, you offered a small $\epsilon$. Per the definition of $sup(A)$, now an $x ~\in A$ must be selected
that satisfies this particular value of $\epsilon$.
Per the definition of $sup(A)$, at no time must an $x ~\in A$ be found that
would satisfy all possible (small positive) values of $\epsilon.$