Proof of the completion of a metric space using cantors diagonal argument and showing a diagonal sequence is cauchy I am studying applied functional analysis out of Applied Analysis by John Hunter. In chp. 1 of the text it gives a proof for the completion of a metric space. I am having trouble with understanding where it actually proves the created metric space is complete. The text can be found here https://www.math.ucdavis.edu/~hunter/book/ch1.pdf and part of the proof I am looking at starts on page 21.
So my questions are: Could someone please help me understand what exactly cantors diagonal argument is and why it would be used here? The only thing I have found so far was that it was another technique that was created to show the reals are uncountable. My second question is why is it important to show that a diagonal sequence is cauchy here? Finally, in the last part of the proof once we have shown the diagonal sequence is a cauchy sequence it goes on to show that $$d(x_{n,k},x_{k,k})\leq \frac{3}{N}$$ as $k\rightarrow\infty$, why do we do this? These questions are solely for deeper understanding and any help and comments would be greatly appreciated. 
 A: First, if you write out your $x_{n,k}$ in a tabular form such as
\begin{eqnarray}
 x_{1,1} & x_{1,2} & x_{1,3} & \ldots \\
 x_{2,1} & x_{2,2} & x_{2,3} & \ldots \\
 \vdots & \vdots & \vdots & \ddots
\end{eqnarray}
Then the diagonal sequence $x_{k,k}$ is being considered. That's the only relation to Cantor's diagonal argument (as you found, the one about uncountability of reals). It is a fairly loose connection that I would say it is not so important. 
Second, $\tilde{X}$, the completion, is a set of Cauchy sequences with respect to the original space $(X,d)$. Why do we need to show that the diagonal sequence $x_{k,k}$ is Cauchy?  Let $x^d$ be the diagonal sequence. Then it would show that $x^d \in \tilde{X}$.  Recall that the goal is to show that $\tilde{X}$ is complete. We started with an arbitrary Cauchy (with respect to $\tilde{d}$) sequence of elements of $\tilde{X}$, and we want to show it converges to something in $\tilde{X}$. This $x^d$ that was constructed is the candidate element, but first we have to show that it is in $\tilde{X}$.
Third, the last part of the proof shows that $\tilde{x}_n$ converges to $x^d$ with respect to the completion metric $\tilde{d}$. If you chase the definition you will see why the book is doing that. That statement the OP references corresponds to the statement that $\tilde{d}(\tilde{x}_n,x^d) < 3/N$ for sufficiently large $n$ (aiming for the definition of convergence under $\tilde{d}$).
Be careful, as there are two levels of "Cauchy" in this proof. We show that $x^d$ is Cauchy with respect to $d$ (to show it is in $\tilde{X}$, as by definition $\tilde{X}$ is a collection of such sequences). The sequence $\tilde{x}_n$ is a Cauchy sequence with respect to $\tilde{d}$, and each element $\tilde{x}_n$ is a sequence itself which is Cauchy with respect to $d$  (A Cauchy sequence of Cauchy sequences... so you already have your hands full with separating these two levels!)
