Prove every Cauchy sequence is a convergent sequence

This is proof every mathematics students will do in their Real Analysis class. In Analysis, you spend a lot of time playing around with inequalities and in this proof you use two inequalities one from the definition and another proof in a clever way to show convergence.

\subsubsection*{Problem 2.} Show that every Cauchy sequence in $\mathbb{R}$ is a convergent sequence in $\mathbb{R}$.
\begin{proof}
Let $\{x_n\} \subset \mathbb{R}$ be a Cauchy sequence.\\
Therefore $\forall \frac{\epsilon}{2} > 0 \ \exists N_0 > 0$ such that $$\forall n,m \geq N_0 \ |x_n-x_m| < \frac{\epsilon}{2}.$$
In addition since $\{x_n\}$ is Cauchy then $\{x_n\}$ is bounded. Since all bounded sequences have a convergent subsequence, there exists a convergent subsequence $\{x_{n_k}\}$, so by definition $\forall \frac{\epsilon}{2} > 0 \ \exists N_1 > 0$ such that $$\forall n_k \geq N_1 \ |x_{n_k} - x| < \frac{\epsilon}{2}.$$
When $n,m,n_k \geq \max\{N_0,N_1\}$ get the following, $$|x_n-x| = |x_n-x_{n_k}+x_{n_k}-x| \leq |x_n-x_{n_k}|+|x_{n_k}-x| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$$
Therefore, $\{x_n\} \subset \mathbb{R}$ is convergent.
\end{proof}

I like this proof because it highlights why the real numbers are so interesting. This proof only holds in complete metric spaces which the real numbers are. And so one way to show a metric space is complete is by showing that every Cauchy sequence converges.