Real Number Line is Complete Metric Space

Theorem
The real numbers $$\R$$, equipped with the usual Euclidean metric, are a complete metric space.

Proof
See Real Number Line is Metric Space for the proof that $$d \left({x, y}\right) = \left|{x - y}\right|$$ is a metric.

It remains to show that this space is complete; i.e. that every Cauchy sequence of real numbers has a limit.

So let $$(a_n)$$ be a Cauchy sequence. It is sufficient to show that $(a_n)$ has a convergent subsequence.

We observe that the fact that $$(a_n)$$ is Cauchy implies that $$(a_n)$$ is bounded.

Indeed, there exists $$n_0 \in \N$$ such that
 * $$ |a_m - a_n| < 1 $$

for all $$m, n \ge n_0$$.

In particular, by the Triangle Inequality:
 * $$ |a_m| = |a_n + a_m - a_n| \le |a_n| + |a_m - a_n| \le |a_n| + 1$$

for all $$m \ge n_0$$.

So the sequence is bounded as claimed.

By the Bolzano-Weierstrass Theorem, $$(a_n)$$ has a convergent subsequence, and we are done.