The reals are complete

For each $k \ge 1$ pick a rational $q_k$ with $|x_k - \hat q_k| < \widehat{1/k}$.

$(q_k)$ is Cauchy in $\mathbb{Q}$: given rational $\varepsilon > 0$, pick $k_0$ with $1/k_0 < \varepsilon/3$ and $K$ with $|x_k - x_l| < \widehat{\varepsilon/3}$ for $k, l \ge K$; then for $k, l \ge \max(k_0, K)$, $\widehat{|q_k - q_l|} \le |\hat q_k - x_k| + |x_k - x_l| + |x_l - \hat q_l| < \widehat{1/k} + \widehat{\varepsilon/3} + \widehat{1/l} \le 3\,\widehat{\varepsilon/3} = \hat\varepsilon$, and the embedding reflects order, so $|q_k - q_l| < \varepsilon$.

Set $x := [(q_k)] \in \mathbb{R}$, the class of this rational Cauchy sequence.

$x_k \to x$: given rational $\varepsilon > 0$, pick $k_1$ with $1/k_1 < \varepsilon/3$ and $K_2$ with $|q_k - q_l| < \varepsilon/3$ for $k, l \ge K_2$; for $k \ge \max(k_1, K_2)$, the difference $\hat q_k - x$ has representative $(q_k - q_l)_l$, whose absolute values $|q_k - q_l|$ are eventually below $\varepsilon/3$, so $|\hat q_k - x| \le \widehat{\varepsilon/3}$, and $|x_k - x| \le |x_k - \hat q_k| + |\hat q_k - x| < \widehat{1/k} + \widehat{\varepsilon/3} \le 2\,\widehat{\varepsilon/3} < \hat\varepsilon$.

Every Cauchy sequence of reals converges in $\mathbb{R}$: the reals are complete.