A sequence is quasi-increasing if: $$\forall \epsilon > 0 \quad \exists N \quad \forall n,m \quad \left( n> m \geq N \implies x_n > x_m - \epsilon \right)$$
My strategy is to show that every bounded quasi-increasing sequence is a Cauchy sequence, i.e: $$ \forall \epsilon>0 \quad \exists N \quad \forall n,m \quad \left( n,m \geq N \implies | x_n - x_m | < \epsilon \right)$$
Since $(x_n)$ is quasi-increasing: $$ \exists K \quad \forall n > m \geq K \quad \left( x_m - x_n < \epsilon/2 \right) $$
Let $s_K = \operatorname{sup} \{ x_n : n \geq K\}$. By definition of supremum: $$ \exists N \geq K \quad x_N > s_K - \epsilon/2 \quad \text{or, equivalently:}\quad s_K - x_N < \epsilon /2 \tag{1} $$
I claim that the sequence is $\epsilon$-Cauchy beyond $N$. Consider any $n,m \geq N$; further, assume that $n > m$. Since $n,m \geq N \geq K$, the sequence is quasi-increasing for $n,m$, therefore: $ x_m - x_n < \epsilon /2 < \epsilon$; this proves the negative side of the absolute difference. For the positive side:
\begin{align} &x_n - x_m \leq s_K - x_m & (\text{$n > N \geq K \implies x_n \leq s_K$})\\ &s_K - x_m < s_K - x_N + \epsilon /2& (\text{$ m \geq N \geq K \implies x_N - x_m < \epsilon /2$})\\ &s_K - x_N + \epsilon /2 < \epsilon /2 + \epsilon / 2& (\text{$s_K - x_N < \epsilon / 2$ from (1)}) \end{align}
For $n < m$, just switch $n$ and $m$. For $n = m$, the difference is $0$, so it always valid. The number $N$ depends on $K$ which depends on $\epsilon$. So there is always a valid $N$ for every $\epsilon$.
