Context: The real numbers were constructed using Cauchy sequences of rational numbers, where every real corresponds to the equivalence class of a rational Cauchy sequence. The fact that $\mathbb{R}$ is Cauchy complete will be used later to prove the Least Upper Bound Property, so we can't use that here.

The following proof was provided for $\mathbb{R}$ being Cauchy complete. Note that $\mathbb{Q}$ being dense in $\mathbb{R}$ has already been proven and the definition of absolute value mentioned in the proof is this one:

Definition. Let $x\in\mathbb{R}$. The \textbf{absolute value} function $|\cdot |:\mathbb{R}\to\mathbb{R}_{\geq 0}$ is defined by \begin{equation*} |x| := \begin{cases} x, & x\geq 0 \\ -x, & x<0 \end{cases} \end{equation*} Furthermore, when considering $x$ as the equivalence class of a Cauchy sequence $\left(x_n\right)$, we may define the absolute value function by \begin{equation*} \left|\left[\left(x_n\right)\right]\right| := \lim_{n\to\infty} \left|x_n\right|. \end{equation*}

Now the proof.

Proof: To prove that $\mathbb{R}$ is Cauchy complete, we wish to show that for all Cauchy sequences $\left(x_n\right)\in\mathbb{R}$, $\exists x\in\mathbb{R}$ such that $\lim_{n\to\infty} x_n = x$. Because $\mathbb{Q}$ is dense in $\mathbb{R}$, it is sufficient to show that for all Cauchy sequences $\left(x_n\right)\in\mathbb{Q}$, $\exists x\in\mathbb{R}$ such that \begin{equation*} \lim_{n\to\infty} \left[\left(x_n,x_n,\ldots\right)\right] = x. \end{equation*} Let the desired $x$ be $x=\left[\left(x_1,x_2,\ldots\right)\right]$. We may observe that $x\in\mathbb{R}$ since $\left(x_n\right)$ is Cauchy sequence of rational numbers (assumed above). Then, by our original statement, we wish to show that \begin{equation*} \lim_{n\to\infty} \left[\left(x_n,x_n,\ldots\right)\right] = \left[\left(x_1,x_2,\ldots\right)\right]. \end{equation*} By definition, this is equivalent to \begin{equation*} \lim_{n\to\infty} \left|\left[\left(x_n,x_n,\ldots\right)\right] - \left[\left(x_1,x_2,\ldots\right)\right]\right| = 0. \end{equation*} Then by the definition of absolute value: \begin{equation*} \lim_{n\to\infty} \left|\left[\left(x_n,x_n,\ldots\right)\right] - \left[\left(x_1,x_2,\ldots\right)\right]\right| = \lim_{n\to\infty} \lim_{m\to\infty} \left|x_n - x_m\right|. \end{equation*} Finally, the limit on the right is $0$ because $\left(x_n\right)$ is Cauchy, which is equivalent to what we originally wanted to show: \begin{equation*} \lim_{n\to\infty} \left[\left(x_n,x_n,\ldots\right)\right] = x. \end{equation*}

My problem: First of all, I'm not clear on the notation. Is $\left(x_n,x_n,\ldots\right)$ an infinite sequence where each term is the $n^{\text{th}}$ term of $\left(x_n\right)$?

Second, if that's the case, why is it sufficient to show that \begin{equation*} \lim_{n\to\infty} \left[\left(x_n,x_n,\ldots\right)\right] = \left[\left(x_n\right)\right]. \end{equation*} (Where $[(x_n)]=[(x_1,x_2,\ldots)]$). Isn't that kind of... self-evident?

Third, I don't understand the premise. Why does $\mathbb{Q}$ being dense in $\mathbb{R}$ $\implies$ it is sufficient to show that for all Cauchy sequences $\left(x_n\right)\in\mathbb{Q}$, $\exists x\in\mathbb{R}$ such that \begin{equation*} \lim_{n\to\infty} \left[\left(x_n,x_n,\ldots\right)\right] = x. \end{equation*} Seeing as how $(x_n)\in\mathbb{Q}$, $[(x_n,x_n,\ldots)]$ is basically just a rational number in $\mathbb{R}$. So, we're saying that the limit of a rational number is equal to a real?

Note: I've looked at this question, but the proofs are completely different, especially since the proof in the above question uses the least upper bound property.

Any help would be much appreciated.