Proving every Cauchy sequence in $\mathbb{R}$ is convergent

Question

I've seen a lot of proofs of how to do this and I don't quite like any of them so I tried proving this myself. Let me know if I am making any logical errors here.

Proof

Let $\varepsilon > 0$ be given.

Since $\{x_n\}_{n=1}^\infty$ is Cauchy, $\forall \varepsilon >0, \exists N \in \mathbb{N} : |x_n-x_m|<\varepsilon$ when $n,m \geq N$

Since $\{x_n\}$ is Cauchy, $\{x_n\}$ is bounded. We know by Bolzano-Weierstrass this means that $\{x_n\}$ has a convergent subsequence $\{x_{n_k}\}$ that converges to $l$.

Put another way: $\forall \varepsilon > 0, \exists N' \in \mathbb{N}: |x_{n_k}-l|<\varepsilon$ when $k>N'$

Suppose $n >$ max$\{N,N'\}$

Then $|x_n-l|=|x_n-x_{n_k}+x_{n_k}-l| \leq |x_n-x_{n_k}|+|x_{n_k}-l| < \varepsilon + \varepsilon = 2\varepsilon.$ $\square$

(Note: I'm not concerned with the fact that my proof ends with $2\varepsilon$. I just wanna know if my statements are true and if my proof logic is good.)

Answer 1

That's fine as long as you have a definition of the real numbers that leads to Bolzano Weierstrass without using Cauchy sequences.