Proof: A convergent Sequence is bounded A part of the proof says that if $n\le N$, then, the sequence 
$x_n \le \max\{|x_1|,|x_2|,....,|x_{N-1}|\}$.

I'm not capturing the intuition of the above. Even more perplexing is, if it the case where the sequence is monotonic decreasing, then for every $n\le N$, it is obvious that $x_n$ will not be $\le \max\{|x_1|,|x_2|,....,|x_{N-1}|\}$.

I greatly appreciate some input.
Well, $|x_n| \le \max\{|x_1|,\ldots, |x_{N-1}|\}$ for all $n \color{red}{&lt;} N$ because if $n &lt; N$, $|x_n|$ is one of the numbers $|x_1|,\ldots, |x_{N-1}|$, and some number in this list is the largest, i.e., $\max\{|x_1|,\ldots, |x_{N-1}|\}$.
If $\{a_n\}$ be a convergent then for any $\epsilon&gt;0$: then there exist a positive integer $n_0$ such that 
$$
|a_n-l|\le \epsilon\:\text{  for all }\:n&gt; n_0
$$
and this implies that $|a_n|\le\epsilon +l$.
Now let $M=\{|a_1|,|a_2|,|a_3|,\ldots,|a_n| ,|\epsilon+l|\}$: then $|a_n|\le M$ is bounded for all $n\in\Bbb N$ and hence sequence $\{a_n\}$ is bounded.