Firstly, let us consider what a completion of a subset of $\mathbb{R}^n$ should be:
Consider some subset $A \subseteq \mathbb{R}^n$, and then define the set
$$\tilde{A}:= A \cup \{\lim_{n \rightarrow \infty} x_n : (x_n)_{n \in \mathbb{N}}\subseteq A\} \subseteq \mathbb{R}^n$$
consisting of all the points in $A$ together with its limit points (where the sequence and limit is considered in $\mathbb{R}^n$). This makes $\tilde{A}$ closed, and thus since $\mathbb{R}^n$ is complete, it makes $\tilde{A}$ complete. In your case this $\tilde{A}$ is the unit ball in $\mathbb{R}^2$.
Since the embedding of $A$ into $\tilde{A}$ should be an isometry, there is only one choice for it, when considering elements of $A$, which $d$ itself.
Since $\tilde{d}$ is necessarily continuous, there is only one choice of metric for elements in $\{\lim_{n \rightarrow \infty} x_n : (x_n)_{n \in \mathbb{N}}\subseteq A\}$; and that is
$$\tilde{d}(\lim_{n \rightarrow \infty} x_n, \lim_{n \rightarrow \infty} y_n) := \lim_{n \rightarrow \infty} d(x_n, y_n).$$
Here it was crucial that there was an ambient space ($\mathbb{R}^n$) to makes this construction; in particular to ensure the existence of the limit points of the sequences $\{\lim_{n \rightarrow \infty} x_n : (x_n)_{n \in \mathbb{N}} \subseteq A\}$.
Now for a more general case: Let some metric space $A$ be given.
The first step is to identify what $\tilde{A}$ should be:
When there is no ambient space, the expressions $\{\lim_{n \rightarrow \infty} x_n : (x_n)_{n \in \mathbb{N}}\subseteq A\}$ are not well defined. However, recall that any convergent sequence is Cauchy. Hence may (naively) define the completion to be the set $A$ together with such elements which we identify as those "to which some Cauchy sequence converges" - i.e. the elements we adjoin are the Cauchy sequences themselves. In order to make this a metric space, the limit of any Cauchy sequence needs to be unique though, so two Cauchy sequences which "converge" to "the same point" should induce the same element. This is where the equivalence relation comes in. Since any constant sequences converges to itself, the $A$ is in bijection with the set of equivalence classes with a constant sequence.
All this amounts to the set
$$ \tilde{A} := \{(x)_{n \in \mathbb{N}} : \text{constant}, x \in A\} \cup \{(x_n)_{n \in \mathbb{N}} : \text{Cauchy}, \forall n \in \mathbb{N}: x_n \in A\} $$
On the first set, the newly defined metric is the same as the original one (which is formally expressed by the map $i$). On the second set, we use the continuity of the metric to define the distance between two adjoint elements.
How should one now think of the completion geometrically?
Visualize where points are that are limit points of sequences in $A$ and then add them to the set.
By the way, a similar (but different) construction is used when defining the real numbers via Dedekind Cuts.
Best Answer
Well, the limits of some of the Cauchy sequences in $X$ may not exist (because $X$ needs not be a complete space), so you need to specify what you mean by "the set of limits". For example, suppose your space consists precisely of points $x_n, n\in\mathbb N$, with $$d(x_n,x_{n+k})=\frac1{2^n}+\frac1{2^{n+1}}+\dots+\frac1{2^{n+k-1}}.$$ There is no limit $p$ to add to $X$.
Of course, you can say "Ah, well, then pick a new point $p$ and declare it to be the limit of the sequence". OK, fine. What is the distance between $x_{17}$ and $p$, in that case?
Now, consider the same example as in the first paragraph, and note that the sequence $x_3,x_6,x_9,\dots$ is Cauchy. Your description says we need to add a new limit point $p'$ corresponding to it. Is $p'\ne p$?
This suggests that the problem is slightly more complicated than anticipated: If two Cauchy sequences are to have the same limit, we better make sure that we add the same limit point for both of them. How do we know that two different sequences ought to have the same limit?
Let's say that we manage to solve all those obstacles, that is: We indeed add to $X$ new points, one for each Cauchy sequence in $X$, making sure that if two distinct Cauchy sequences are to converge to the same thing, they indeed do. We also somehow manage to extend the distance function so the new space is metric, and $X$ is dense in it. How do we know that there is not a Cauchy sequence of new points for which we haven't yet added a limit point? Because if there are such sequences, then we ought to iterate the procedure. For how long? Does it ever end? Now, if there are no such sequences, that is certainly part of what the proof needs to show.
All that being said, you are on the right track. First we need to deal with what things we are to add. Of course, new points, and the new points can be anything we want, but let's try to choose something specific so we can keep track of them. A natural thing to do is to exploit the fact that if two Cauchy sequences are to have the same limit, then we better add the same point as limit of both of them. One way to take advantage of this is to introduce an equivalence relation on Cauchy sequences, saying that two such sequences are equivalent "if they are to have the same limit". Then as the new points we can just add the equivalence classes of this equivalence relation.
How do we check that two distinct sequences have the same limit? Luckily, this is easy: Say the sequences are $x_1,x_2,\dots$ and $y_1,y_2,\dots$ Define a new sequence by $$z_1=x_1,z_2=y_1,z_3=x_2,z_4=y_2,z_5=x_3,\dots$$ Then the two sequences are equivalent iff the new sequence so described is Cauchy. It does not matter here whether there are repetitions in this sequence of $z_i$. (Naturally, there are things to verify here, mainly that this is indeed an equivalence relation.)
A small problem at this point is that some sequences may be Cauchy and already have a limit in $X$. The easiest solution is to ignore them. It may not be the prettiest solution, because now your space consists of two rather different creatures: Elements of $X$, and equivalence classes of Cauchy sequences of elements of $X$, that do not converge to an element of $X$. But it is a fine solution (meaning: It works). The standard approach is to avoid this separation of creatures, and simply take as the new space the collection of all equivalence classes of Cauchy sequences. (If a sequence converges to $x\in X$, we identify its equivalence class with $x$, so rather than the isometry being inclusion, at the end we have something a tad more elaborate.)
Now comes the second problem: How do we make this thing into a metric space? Luckily, there is an easy solution as well: If $x_n\to p$, then for any $k$, we have that $d(x_k,x_n)\to_{n\to\infty}d(x_k,p)$. So we can use this as the way to define the new distances: Given an equivalence class $p$, let $x\in X$. We define $d(x,p)$ as $\lim_{n\to\infty}d(x,x_n)$, where $ x_1,x_2,\dots$ is some Cauchy sequence in the equivalence class $p$. OK. Maybe not so easy: We need to check that this definition gives us a positive real number (as opposed to $0$ [excluded since the $x_n$ do not converge in $X$, so $x$ better not be their limit], or $+\infty$, or to the case when the limit does not exist). We also need to check that this number is independent of the sequence $x_1,x_2,\dots$ we picked. A similar idea gives us how to define $d(p,q)$ when both $p,q$ are equivalence classes.
Of course, one still needs to verify this indeed gives us a metric space.
Finally, we need to check that this is complete, and $X$ is dense in it. But I'll stop here, as I'm pretty sure the construction in your book is following the same lines.
(There are rather different presentations of the construction, that look superficially different and start from very different ideas, but the space obtained at the end is essentially unique, in the sense that any two constructions will be isometric via an isomorphism that identifies their copies of $X$. Naturally, this also takes an argument.)