I want to show that Cauchy sequences are bounded without using the fact that Cauchy sequences are convergent sequences and convergent sequences are bounded.
proof: Suppose $a_m$ is a Cauchy sequence, that is $\forall\varepsilon\gt0, \exists N\in\mathbb{R}$ s.t.
$$m,n\gt N\implies |a_m-a_n|\lt \varepsilon$$
So we have $$-\varepsilon\lt a_m-a_n\lt\varepsilon$$ and $$|a_n|-|\varepsilon|\le a_n-\varepsilon\lt a_m\lt a_n+\varepsilon \lt |a_n|+|\varepsilon|$$ $$|a_m|\lt |a_n|+\varepsilon$$
Am I on the right track at all? Can I then say that $M$ will be the bound for $a_m$ as: $$M=\max\{a_1,...,|a_{n-1}|,|a_n|+\varepsilon\}$$Or am I making some incorrect assumption or something else?
I would suggest that you break the sequence up into a finite sequence and the remaining infinite sequence. So, fill in the details in the following proof.
Proof: Given that $(a_n)_{n=1}^\infty$ is a Cauchy sequence, for $\varepsilon =1$, there exists an $N$ such that for all $j,k>N$, we have that $|a_j-a_k|<1$
Now split the sequence into a finite sequence $(a_n)_{n=1}^N$ and the remaining infinite sequence $(a_n)_{n=N+1}^\infty$. Now use the theorem(use induction to prove it) that every finite sequence is bounded. So there exists a real number $M$ such that the sequence $(a_n)_{n=1}^N$ is bounded by $M$.
Now for the sequence $(a_n)_{n=N+1}^\infty$, we have the inequality that for every $j>N$ and fixed $k>N$ we have that $|a_j-a_k|<1$.
Take it from here to conclude that $|a_j|<1+|a_k|$. Then use your previous result that $(a_n)_{n=1}^N$ is bounded by $M$ to conclude that the entire sequences $(a_n)_{n=1}^\infty$ is bounded by $1+|a_k|+M$.