I am self-studying the book Understanding Analysis by S. Abbott and I tried to solve the exercise 2.6.6 (c) without relying on Bolzano-Weierstrass Theorem as they did here. Such exercise requires the following definition.
Lets call a sequence quasi-incrasing if for all $\varepsilon>0$ there exists an $N$ such that whenever $n>m\geq N$ it follows that $a_n>a_m-\varepsilon.$
Now, I want to prove that if $(a_n)$ is a quasi-increasing and bounded sequence, then $(a_n)$ must converge. My attempt of proof goes as follows.
Suppose, in order to derive a contradiction, that $(a_n)$ does not converge. Then by the Cauchy Criterion, $(a_n)$ is not a Cauchy sequence. Hence, there must be an $\varepsilon_0$ such that for all $N \in \mathbb{N}$ there exists $n,m \geq N$ such that $|a_m-a_n|\geq \varepsilon_0.$ (*)
On the other hand, as $(a_n)$ is a quasi-increasing sequence there is an $N \in \mathbb{N}$ such that for all $n>m\geq N$, it follows that $a_n>a_m-\frac{\varepsilon_0}{2}.$ (**)
Also, the boundedness of $(a_n)$ guarantees the existence of an $M>0$ such that $|a_n| \leq M$ for all $n \in \mathbb{N}.$
Now, by (*) we can choose $n,m \geq N$ such that $|a_{n}-a_{m}| \geq \varepsilon_0.$ WLOG, assume $n>m$ (observe that it cannot happen that $n=m$). Then we have the inequality $a_n>a_m-\frac{\varepsilon_0}{2}.$
Now, since $|a_m-a_n|\geq \varepsilon_0>\frac{\varepsilon_0}{2}$ and $\frac{\varepsilon_0}{2}>a_m-a_n,$ then it must be satisfied the inequality $$a_n>a_m+\frac{\varepsilon_0}{2}.$$
Choose $n_1=m$ and $n_2=n.$
Suppose that we have choosen $n_1<\cdots<n_k$ such that for all $j=1,\ldots,k-1$ it follows the inequality $$a_{n_{j+1}}>a_{n_j}+\frac{\varepsilon_0}{2}.$$
Now choose $N'=n_k+1.$ Observe that $N'>n_k>n_1> N.$ Again, by (*) we can choose $n',m'\geq N'$ such that $|a_{m'}-a_{n'}| \geq \varepsilon_0.$ WLOG assume $n'>m'.$
We have now the two following conclussions:
(1) Since we have $n'>m'\geq N'>N$, then by (**) we must have $a_{n'}>a_{m'}-\frac{\varepsilon_0}{2}>a_{m'}-\varepsilon_0,$ and hence $\varepsilon_0>a_{m'}-a_{n'}.$ This, together with the fact that $|a_{m'}-a_{n'}| \geq \varepsilon_0$ allows us to conclude that $$a_{n'}\geq a_{m'}+\varepsilon_0.$$
(2) Since $m'\geq N'>n_k>N,$ then by (**) we have that $a_{m'}>a_{n_k}-\frac{\varepsilon_0}{2}.$
Then, by (1) and (2) we should obtain $$a_{n'} \geq a_{m'}+\varepsilon_0>\left(a_{n_k}-\frac{\varepsilon_0}{2}\right)+\varepsilon_0)=a_{n_k}+\frac{\varepsilon_0}{2}.$$
Choose now $n_{k+1}=n'>N'>n_k.$
We have then constructed inductively a subsequence $(a_{n_k})$ of $(a_n)$ that satisfies $$a_{n_{k+1}}>a_{n_k}+\frac{\varepsilon_0}{2} \quad \text{(***)}$$ for all $k \in \mathbb{N}$.
Lastly, choose a number $K \in \mathbb{N}$ such that $K\frac{\varepsilon_0}{2}\geq M-a_{n_1}.$ So, by the condition (***) applied several times we obtain $$a_{n_{K+1}}>a_{n_K}+\frac{\varepsilon_0}{2}> a_{n_{K-1}}+2\frac{\varepsilon_0}{2}> \cdots > a_{n_2}+(K-1)\frac{\varepsilon_0}{2} >a_{n_1}+K\frac{\varepsilon_0}{2}\geq M.$$
This contradicts the fact that $|a_{n_{K+1}}|\leq M.$
Then we have the desired conclussion.
I hope you can check this proof and let me now if there is a mistake. Thanks in advance.
