Please check my proof
Suppose that sequence is bounded then it exist $a_{n}< M$
Since it is monotone increase sequence and bouned, then limit exist
By defination if limit of sequence it will exist $|a_{n}-L|<\epsilon $
that it shows it cauchy sequence.
What is the definition of Cauchy Sequence you are using? You are using two theorems here, first that every bounded monotone sequence has a limit, and second that every convergent sequence is Cauchy - that seems heavy duty to me.
Using the fact that every set of real numbers which is bounded above has a least upper bound, we can choose $M$ as the least upper bound of the sequence.
Since $M$ is the least upper bound we know that $M-\epsilon$ is not an upper bound, so there is an $N$ with $M-\epsilon \lt a_N \le M$. Now suppose that $m\gt n\gt N$. The sequence is monotone increasing, so $a_m\ge a_n\ge a_N$ (not assuming strictly increasing). We therefore have $$M-\epsilon \lt a_N\le a_n\le a_m \le M$$ from which it follows that $0\le a_m-a_n \lt \epsilon$. So we can demonstrate that the Cauchy condition holds.
It is, of course, equally easy to prove that the limit is $M$, but this is not what is asked. My instinct is that you are intended to show that you know what the Cauchy condition is, and show explicitly that it holds.