Convergence Of Cauchy's Sequencel Prove that a Cauchy's Sequence always converges.
$$$$Consider a Cauchy's Sequence $a_n$ then it has the property that for every $\epsilon >0$ there exists a $M$ such that whenever $m, n \geq M$ we have $$|a_n-a_m|<\epsilon$$. Now choose a $\epsilon_0$, then there exists a $M_0$ such that whenever $m, n \geq M_0$ we have $$|a_n-a_m|<\epsilon_0$$. Now choose any $m_0>M_0$ then for all $n>M_0$ we have $$a_{m_0}-\epsilon_0<a_n<a_{m_0}+\epsilon_0$$ and for all $n<M_0$ we can choose the maximum and minimum value among all $a_n$. So we have $a_n$ is bounded. So there exists an increasing or decreasing subsequence of $a_n$ say $b_n$ which converges. Now WLOG let us assume that $b_n$ is increasing. Now as $b_n$ is bounded so let $L$ be it's least Upper bound so $b_n$ converges to $L$.So for every $\epsilon>0$ there exists a $N$ such that for all $n>N$ we have $$|b_n-L|<\epsilon$$. Now as $b_n$ is a subsequence of $a_n$ so for every $r$ there exists a $s>r$ such that $b_r=a_s$ and hence whenever $n>M$ $b_n$ satisfies $$|b_n-a_m|<\epsilon$$for all $m>M$. Now WLOG let us assume that $\text{max}(N, M)=N$.Now from the definition we have for every $\epsilon$ there exists a $N$ such that whenever $m, n>N$ we have $$|b_n-a_m|<\epsilon$$. So for every $N$ we can choose a $n_N>N$ then for every epsilon there exists a $n_N$ such that whenever $m>n_N$ we have $$|b_n-a_m|<\epsilon$$ and hence we have for every $\epsilon>0$ there exists a $n_N$ such that for all $m>n_N$ we have $$|a_m-L|=|a_m-b_n+b_n-L| \leq |a_m-b_n|+|b_n-L| <2\epsilon$$ and as $\epsilon$ is arbitrary so we have $a_n$ converges to the limit $L$
$$$$Is My Proof Correct????
 A: The part where you say "has an increasing or decreasing subsequence which converges" holds only when ypu are working in a complete metric space, but that is not mentioned here, which will lead to the problem being wrong. The converse is true always, but for this implication, you need to have the first condition.
A: In context:
1) Completeness axiom.
In $\mathbb{R}$  every Cauchy sequence converges.
2) Bolzano Weierstrass: 
Every bounded sequence of real numbers has a convergent subsequence.
To prove Bolzano Weiertrass one uses the completeness of $\mathbb{R}$.
Your proof:
By using Bolzano Weierstrass you use the completeness of $\mathbb{R}$, i.e. the existence of a limit $L$ of a subsequence.
Then you show that this $L$, given that $a_n$ is Cauchy, is the limit of $a_n$.
Summary: 
1) Completeness of $\mathbb{R}$: If $a_n$ is Cauchy then it has a limit $L$.
2) Bolzano Weierstrass: You use the limit $L$ of the subsequence, and $a_n$ is Cauchy, to show  $a_n$ converges to this limit.
3) The cardinal point is the completeness of $\mathbb{R}$.
