Completion of metric spaces (explanation of the proof) In the book Introductory Functional Analysis with Applications (Kreyszig), the author proves the following theorem:

For a metric space $X = (X, d)$ there exists a complete metric space $\hat X=(\hat X, \hat d)$ which has a subspace $W$ that is isometric with $X$ and is dense in $\hat X$. This space $\hat X$ is unique except for isometries, that is, if $\tilde X$ is any complete metric space having a dense subspace $\tilde W$ isometric with $X$, then $\hat X$ and $\tilde X$ are isometric. (page 41)

The metric in $\hat X$ is, by definition, $\hat d(\hat x,\hat y)=\lim(x_n,y_n)$ (see page 42 for details). The last step is to prove that $\hat X$ is unique except for isometries. So, is supposed that there exists other complete metric space $(\tilde X,\tilde d)$ with a subspace $\tilde W$ dense in $\tilde X$ and isometric with $X$. (page 45)
My question is: in this proof, should $\tilde d$ be arbitrary? If so, why is stated $\tilde d(\tilde x, \tilde y)=\lim\tilde d(\tilde x_n,\tilde y_n)$? Seems that a particular metric is used.
Thanks.
 A: The metric $\tilde d$ cannot be completely arbitrary: $\langle\tilde X,\tilde d\rangle$ is required to be a complete metric space with a dense subspace $\tilde W$ isometric with $X$. Within those requirements, however, it is arbitrary. Let $\tilde x,\tilde y\in\tilde X$ be arbitrary. Since $\tilde W$ is dense in $\tilde X$, there are sequences $\langle \tilde x_n:n\in\Bbb N\rangle$ and $\langle \tilde y_n:n\in\Bbb N\rangle$ in $\tilde W$ converging to $\tilde x$ and $\tilde y$, respectively. Let $\epsilon>0$ be arbitrary. There are $m_x,m_y\in\Bbb N$ such that $\tilde d(\tilde x_n,\tilde x)<\frac{\epsilon}2$ for all $n\ge m_x$ and $\tilde d(\tilde y_n,\tilde y)<\frac{\epsilon}2$ for all $n\ge m_y$. Let $m=\max\{m_x,m_y\}$; then
$$\begin{align*}
\tilde d(\tilde x,\tilde y)&\le\tilde d(\tilde x,\tilde x_n)+\tilde d(\tilde x_n,\tilde y_n)+\tilde d(\tilde y_n\tilde y)\\
&<\frac{\epsilon}2+\tilde d(\tilde x_n,\tilde y_n)+\frac{\epsilon}2\\
&=\tilde d(\tilde x_n,\tilde y_n)+\epsilon
\end{align*}$$
and
$$\begin{align*}
\tilde d(\tilde x_n,\tilde y_n)&\le\tilde d(\tilde x_n,\tilde x)+\tilde d(\tilde x,\tilde y)+\tilde d(\tilde y,\tilde y_n)\\
&<\frac{\epsilon}2+\tilde d(\tilde x,\tilde y)+\frac{\epsilon}2\\
&=\tilde d(\tilde x,\tilde y)+\epsilon
\end{align*}$$
for all $n\ge m$, so $\left|\tilde d(\tilde x,\tilde y)-\tilde d(\tilde x_n,\tilde y_n)\right|<\epsilon$ for all $n\ge m$. It follows that $$\lim_{n\to\infty}\tilde d(\tilde x_n,\tilde y_n)=\tilde d(\tilde x,\tilde y)\;.$$
This is just a consequence of the properties of a metric and the fact that the two sequences converge to $\tilde x$ and $\tilde y$, respectively. $\langle\tilde X,\tilde d\rangle$ could be any metric space, and $\langle\tilde x_n:n\in\Bbb N\rangle$ and $\langle\tilde y_n:n\in\Bbb N\rangle$ any sequences in $\tilde X$ converging to $\tilde x$ and $\tilde y$, respectively, and you’d have $$\lim_{n\to\infty}\tilde d(\tilde x_n,\tilde y_n)=\tilde d(\tilde x,\tilde y)\;.$$
A: They first construct a space with the properties. They give this space a metric given by $\hat{d}(\hat{x},\hat{y}):=\lim_n d(x_n,y_n)$ (Notice the $d$ inside the limit is the metric in the original metric space before completion). It is a specific metric. 
Then they proceed to prove that any other metric space with the properties in the definition is isometric to this one.
