Ross Elementary Analysis Theorem 11.2 proof

The following is part of a proof for theorem 11.2 in Ross's Elementary Analysis.

Let $(s_n)$ be a sequence and $t \in \mathbb{R}$. In the proof, we assume that\begin{equation}\{n \in \mathbb{N} : t - \epsilon < s_n < t \} \text{is infinite for all $\epsilon > 0$.}\tag{1}\label{1} \end{equation}Now we are trying to prove by construction that a subsequence with a limit exists. We select $n_1$ such that $t-1 < s_{n_1} < t$. Then we suppose that $n_1 < n_2 < \ldots < n_{k-1}$, and\begin{equation}\max \{s_{n_{j-1}}, t - \frac{1}{j}\} \leq s_{n_j} < t \quad \text{for $j = 2, \ldots, k-1$}\tag{2}\label{2}\end{equation}

The proof states that if we use (\ref{1}) by setting $\epsilon = \max \{s_{n_{k-1}}, t - \frac{1}{k}\}$, we can select $n_k > n_{k-1}$ satisfying equation (\ref{2}).

I am unsure how this choice of $\epsilon$ satisfies equation (\ref{2}). Could someone help explain this?