If $i[X]$ where not dense in $(\tilde{X}, \tilde{d})$, pick $p$ in $\tilde{X}$ not in the closure of $i[X]$. We know $r>0$ exists so that $B_{\tilde{d}}(p,r) \cap i[X] = \emptyset$.
Define $Y=\overline{i[X]}$ in the restricted metric $d_Y$ from $\tilde{d}$, and note it is a complete metric space (being closed in one) and $f=i$ is well-defined isometry from $(X,d)$ to $(Y,d_Y)$. Now suppose an isometry $g: (\tilde{X}, \tilde{d}) \to (Y,d_Y)$ completing the diagram (so $g \circ i = f$) existed, then where can $p$ go?
$g(p) \in Y$ so lies (in $\tilde{X}$) in the closure of $i[X]$, so for some $x_0 \in X$, $\tilde{d}(i(x_0), g(p)) < r$.
But $i(x_0)= f(x_0) = g(i(x_0))$ and then $d_Y(g(p), g(i(x_0)) < r$ while we knew $\tilde{d}(p, i(x_0)) \ge r$, so $g$ is not an isometry.
So $i[X]$ has to be dense in $\tilde{X}$.
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
Take as our starting metric space $[0,1)$. Obviously the "right" completion is $[0,1]$, that is:
take the original space;
add a new point to it, namely $1$;
extend the metric to set $d(a,1)=1-a$ for each $a\in [0,1)$.
However, this isn't the only thing we could do. For example, we could also look at the metric space whose underlying set is $[0,1)\cup\{17\}$ with distance function $f$ given by $f(x,y)=\vert x-y\vert$ if $x,y\in [0,1)$, $f(17,17)=0$, and for $a\in [0,1)$ we set $f(a,17)=f(17,a)=1-a$.
This sort of issue indicates why "definitions" like "the smallest complete metric space containing $R$" are going to be problematic: they require us to compare objects which don't, on the face of it, admit any sort of meaningful comparison. That said, there are a couple ways to make your proposed definition precise, which I'll phrase as theorems (using the standard definition of "completion"):
Suppose $R\subseteq R^*$ are metric spaces with $R^*$ complete. Then $R^*$ is a completion of $R$ iff no proper subspace of $R^*$ containing $R$ is complete. (So completions are "internally minimal.")
Suppose $R\subseteq R^*$ are metric spaces with $R^*$ complete. Then $R^*$ is a completion of $R$ iff for every complete metric space $S$ with $R\subseteq S$ there is a unique isometric embedding of $R^*$ into $S$ which is the identity on $R$. (So completions are "minimal with respect to comparisons via isometries.")
The second notion isn't as snappy, but it's actually getting at a deeper idea than the first (that of figuring out how to compare very disparate objects by looking for maps between them, especially unique maps, satisfying certain nice properties).