(1) $\epsilon$-$\delta$ Criterion: For each $\epsilon > 0$, there is a $\delta > 0$ such that, for all $x$ in $\text{Dom}(f)$, $|x - c| \leq \delta \implies |f(x) - f(c)| \leq \epsilon$.
In this book's proof that continuity implies the above criterion, we have:
... we give a contrapositive proof. That is, assume there exists an $\epsilon > 0$ so that there is no $\delta$ so that (1) holds. We'll show that $f$ is not contnuous at $c$. Let $\delta_n = \frac{1}{n}$. Since (1) does not hold, there is an $x_n \in \text{Dom}(f)$ satisfying $$|x_n - c| \leq \delta_n$$ such that $$|f(x_n) - f(c)| > \epsilon$$
I bolded the statement I'm having an issue with: Why can we make this statement? We know that, for some $\epsilon$, (1) does not hold, but where does that allow us to assume a particular $\delta_n$? Where were we able to assume that that $|x_n - c| \leq \delta_n$ for some $x_n$?
I know it may seem like somewhat of a trivial question, but I do not see where this assumption is validated.