Example of proof of an infimum?

I'm doing an excercise about infimums and supremums and I've seen different examples of proving a =inf(S) $\iff \forall \epsilon >0\exists s\in S\colon a+\epsilon>s$ and to me they just seem to be making it up like:

Let S := { $x+1, x\in (0,1)$}

Here the infimum is clearly 1 , so let's prove the hypothesis inf(S) = 1. Which means $1+\epsilon >x+1 $ for some $x \in (0,1)$.

Then it usually goes as, solving for x as scratchwork such that if $$ 1+\epsilon < x_n +1 \implies x_n> \epsilon $$ Now, going backwards:

Given$$x_n>\epsilon \implies x_n +1 > \epsilon +1 $$

Then it is proven. However I fail to see how doing the same scratchwork for a different minimum for example $-2$

Scratchwork:

$$-2 +\epsilon < x_n +1 \implies x_n > \epsilon -3 $$

""Proof"":

Given $x_n > \epsilon -3 \implies x_n +1 > \epsilon -2 $

I believe my problem is with the bounds of x, since in the "proof" epsilon has to be at least 3 for a valid x to exist. But then I don't quite know how to express the $\forall\epsilon > 0 $ because in the correct proof the inequality is not valid for $\epsilon > 1 $