Prove that if a sequence is a Cauchy sequence, then all of its subsequences converge. Suppose that we do not know that every real Cauchy sequence converges. Since we know that if every subsequential limit of a sequence of real numbers converges to the same number, then the sequence itself converges to that number, let us try to get from a Cauchy sequence to this result.
Question: Suppose $(x_n)$ is a Cauchy sequence of real numbers. Show that every subsequential limit converges to the same real number.
I have been struggling to come up with a solid proof for this question. I keep getting stuck after applying the fact that a Cauchy sequence is bounded and therefore has a convergent subsequence. How in the world do I show that all the subsequences converge to the same number?
Thank you in advance.
 A: Part $1$: Convergent subsequence
From the definition of Cauchy Sequence, there exists $N\in\mathbb{N}$ such that $n\geq N$ implies $|x_N-x_n|<1$. This then implies that for all $n\geq N$ we have
$$x_N-1<x_n<x_N+1$$
and therefore for all $n\in\mathbb{N}$
$$\min\{x_1,x_2,...,x_{N-1},x_N-1\}\leq x_n \leq \max\{x_1,x_2,...,x_{N-1},x_N+1\}$$
This implies that $x_n$ is a bounded sequence. Then by the Bolzano–Weierstrass theorem we may conclude that $x_n$ has at least one convergent subsequence. Let us denote this subsequence by $x_{n_k}$ and let its limit be
$$\lim_{k\to\infty} x_{n_k}=L$$
Part $2$: Entire sequence converges
Now, let $\epsilon>0$ be given. By definition, there exists $N_1\in\mathbb{N}$ such that $k\geq N$ implies $|x_{n_k}-L|<\frac{\epsilon}{2}$. Additionally, there exists $N_2\in\mathbb{N}$ such that $N_2\leq n\leq m$ implies $|x_n-x_m|<\frac{\epsilon}{2}$. Now, choose $N=\max\{n_{N_1},N_2\}$ and let $l$ be the smallest natural such that $n_l\geq N$ and $x_{n_l}$ is a member of the convergent subsequence. Then for all $n\geq N$ we have
$$\bigg||x_n-L|-|x_{n_l}-L|\bigg|\leq |(x_n-L)-(x_{n_l}-L)|=|x_n-x_{n_l}|<\frac{\epsilon}{2}$$
(the first step in this equation is the reverse triangle inequality). This then implies
$$0\leq  |x_n-L|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$
(since $|x_{n_l}-L|<\frac{\epsilon}{2}$). We conclude that
$$\lim_{n\to\infty}x_n=L$$
Part 3: Convergent subsequences
Let $\epsilon>0$ be given. By defintion there exists $N\in\mathbb{N}$ such that $n\geq N$ implies $|x_n-L|<\epsilon$. Now, let $x_{n_k}$ be any subsequence of $x_n$ and note that $n_k\geq k$. Therefore, for all $k\geq N$ we have $n_k\geq k\geq N$. This implies
$$|x_{n_k}-L|<\epsilon$$
We conclude
$$\lim_{k\to\infty}x_{n_k}=L$$
