Can someone help me understand the proof that every cauchy sequence is bounded? This proof is written by a user Batman as an answer to someone's question(just to give credit). Every proof that I've seen is the same idea, and I'm having trouble understanding it intuitively. (I don't see why to take n=N and why the max is used.)  If someone could explain it in words it would be really helpful.
Choose ϵ>0. Then, there exists N such that for m,n ≥ N, |am−an|<ϵ. By the triangle inequality, |am|−|an|≤|am−an|<ϵ. Take n=N and we see |am|−|aN|<ϵ for all m≥N.
Rearranging, we have |am|<ϵ+|aN| for all m≥N. Thus, |am|≤max{|a0|,|a1|,…,|aN1|,ϵ+|aN|} for all m. Thus, am is bounded (it is sandwiched in ±max{|a0|,|a1|,…,|aN1|,ϵ+|aN|}).

Just to make things simpler, let's take $\epsilon=1$.  Since the sequence is Cauchy, there is an integer N such that $m,n\ge N\implies|a_m-a_n|<1$.  In particular, taking $n=N$, we have that $|a_m-a_N|<1$ for $m\ge N$;
so $|a_m|=|(a_m-a_N)+a_N|\le |a_m-a_N|+|a_N|<1+|a_N|$ for $m\ge N$ using the Triangle inequality.
Now we have a bound for all but finitely many of the terms of the sequence, but we still have to account for the terms $a_1,a_2,\cdots,a_{N-1}$.
If we take $M=\max\{|a_1|,\cdots,|a_{N-1}|, 1+|a_N|\}$, then $|a_n|\le M$ for all n
since $|a_n|\le M$ if $m\ge N$ by the first part, and $|a_n|\le M$ for $1\le m\le N-1$ by the definition of M.
