In Abbott, Understanding Analysis page 17 we have the following:
Lemma 1.3.8. Assume $s \in \mathbb R$ is an upper bound for a set $A \subseteq \mathbb R$. Then, $s = \sup A$ if and only if, for every choice of $\epsilon > 0$, there exists an element $a \in A$ satisfying $s-\epsilon < a$.
But if we take $A = [1, 2]$ then $s = \sup A =2$ and choose $\epsilon = 5 > 0$, say, then the lemma fails since $s-\epsilon = 2-5=-3 \notin A$?
$s-\epsilon$ need not be in $A$. The Lemma says that the is an element of $A$ that is bigger than $s-\epsilon$, although $s$ is bigger than (or equal) to all elements of $A$.
In your example, where we take $\epsilon=5$ all the elements of $A$ are bigger than $s-\epsilon=2-5=-3$. In fact, for every choice of $\epsilon>0$, the element $2\in A$ is bigger than $s-\epsilon=2-\epsilon$.
Every number $s'$ greater than 2 cannot be the supremum, since we can take $\epsilon=\frac{s'-2}{2}$ so $s'-\epsilon=\frac{2+s'}{2}>2$, so for this choice of $\epsilon$, all the elements of $A$ are less than $s'-\epsilon$.