How can the completeness of $\mathbb{R}$ (every Cauchy sequence in $\mathbb{R}$ converges) be proven from the Bolzano-Weierstrass Theorem?

We can also use the following theorems as the ingredients:
1. Every sequence in $\mathbb{R}$ has a monotone subsequence.
2. Every bounded monotone sequence in $\mathbb{R}$ converges.
3. Every Cauchy sequence in any metric space is bounded.

This is how I tried:

Theorem 1 and Theorem 3 imply that every Cauchy sequence in $\mathbb{R}$ is bounded and has a monotone subsequence. We also know that a subsequence of a bounded sequence is bounded. Hence we get every Cauchy sequence in $\mathbb{R}$ is bounded and has a bounded monotone subsequence.

Theorem 2 implies that the bounded monotone subsequence of the Cauchy sequence mentioned previously has a subsequence that converges.

This is the part where I got stuck. It seems that I keep getting subsequences from a subsequence and only manage to show that the subsequence converges, not the original Cauchy sequence.

Can anybody show where did I do wrong and how can I fix it?

Thanks for the help!


The key point is that in any metric space, if a Cauchy sequence has a convergent subsequence, then the original sequence converges as well. Indeed, suppose $(x_n)$ is Cauchy and the subsequence $(x_{n_k})$ converges to some $x$. Then for any $\epsilon>0$, there is an $N$ such that $d(x_{n_k},x)<\epsilon$ whenever $n_k>N$ and such that $d(x_m,x_n)<\epsilon$ whenever $m,n>N$. For any $n>N$, we can then choose $k$ so that $n_k>N$, and we get $d(x_n,x)\leq d(x_n,x_{n_k})+d(x_{n_k},x)<2\epsilon$.

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