Alternative approach to prove that if a sequence is a Cauchy Sequence, then it converges. I am familiar with the traditional proof that if a sequence of real numbers is a Cauchy sequence, then it is convergent. However, I am stuck on a problem that asks me to prove the implication using the fact that if all the subsequences of a sequence converge to a real number then the sequence itself converges.
Question: (i) Let $(a_n)$ be a bounded sequence of real numbers and let $x \in \mathbb{R}$ have the property that every subsequence of $(a_n) \neq (a_n)$ converges to $x$. Show that $(a_n) \to x$. (ii) Furthermore, use this result to prove that if a sequence is a Cauchy sequence, then it converges.
Proving part (i) was fairly easy for me. Outlining a proof for part (ii) is tripping me up. I am having a hard time applying the properties of a Cauchy sequence to the proof other than the fact that this implies the sequence is bounded.
Any help is greatly appreciated.
 A: For (i), simply let $b_n:=a_{n+1} ,\forall n\ge1.$ Then by the assumption, $(b_n)$ is a convergent subsequence of $(a_n)$ , and it implies obviously that $(a_n)$ is a convergent sequence by the definition of limits.
For (ii), try to argue by contradiction with the definitions of convergent sequences, limits, and Cauchy sequences.

 The basic idea is that, if a sequence $(a_n)$ is divergent, then no matter how rear in the "tail"(i.e. $\{a_n\,\vert n\ge k, k$ large$\}$, there always exist two elements of it which are not close enough. However, every two terms with sufficiently large indexes of a Cauchy sequence can get sufficiently close. This is the contradiction.

A: Background of this question: Some lecturers in analysis use this method to prove that Cauchy implies convergence. Generally, these lecturers teach it after teaching Bolzano-Weierstrass Theorem. Now let's prove it.
Suppose $\{a_j\}$ is Cauchy, then we know $\{a_j\}$ is bounded. (Recall that every Cauchy sequence is bounded.) By B-W theorem, there exists a subsequence $\{a_{j_k}\}$ which converges to $c$. Notation: $\lim\limits_{k \to \infty} a_{j_k}=c$
Our goal is to show $\lim\limits_{j \to \infty} a_{j}=c$
$\forall \epsilon >0$, we need to show $\exists N \in \mathbb{N}$, s.t. whenever $j>N$, we have $|a_j-c|<\epsilon$.
But we know $\{a_j\}$ is Cauchy, thus given$\frac{\epsilon}{2}>0$, $\exists N \in \mathbb{N}$ s.t. whenever $j>k>N$, we have $|a_j-a_k|<\frac{\epsilon}{2}$.
Now when $j>N$, $|a_j-c| \leq |a_j-a_{j_k}+a_{j_k}-c|\leq |a_j-a_{j_k}|+|a_{j_k}-c|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$, where we choose $k$ so big s.t. $j_k > N$ and $|a_{j_k} - c|<\frac{\epsilon}{2}$.
