Compactification of separable metric spaces Suppose $(X,d)$ is a separable metric space with $d\leq 1$, and let $(x_n)$ and $(x'_n)$ be two dense sequences. Define embeddings
$$\varphi,\varphi':X\to[0,1]^\mathbb{N}$$
by putting $\varphi(x)=(d(x,x_n))_{n}$ and $\varphi'(x)=(d(x,x'_n))_{n}$ respectively. Then $\mathcal{X}=\overline{\varphi(X)}$ and $\mathcal{X}'=\overline{\varphi'(X)}$ are two metrizable compactifications of $X$.
It seems to me that $\mathcal{X}$ and $\mathcal{X}'$ always homeomorphic. In order to define a maps $\mathcal{X}\xrightarrow\theta\mathcal{X}'$ and $\mathcal{X}'\xrightarrow{\theta'}\mathcal{X}$ we define $\theta$ on $\varphi(X)$ as $\varphi'\circ\varphi^{-1}$ (I'm cutting corners here), and extend it by uniform continuity (and similarly for $\theta'$). I haven't checked the details, but this will produce (uniformly) continuous maps $\mathcal{X}\xrightarrow\theta\mathcal{X}'$ and
$\mathcal{X}'\xrightarrow{\theta'}\mathcal{X}$ satisfying $\theta\circ\varphi=\varphi'$ and $\theta'\circ\varphi'=\varphi$ so that $\theta\circ\theta'$ coïncides with the identity function on the dense subset $\varphi(X)\subset\mathcal{X}$, and $\theta'\circ\theta$ coïncides with the identity function on the dense subset $\varphi'(X)\subset\mathcal{X}'$. And thus we should get that $\theta$ and $\theta'$ are inverse homeomorphisms of one another.

Question : Is the above reasoning sound, and if so, what is the name of this compactification ? Does it have a universal property ? Maybe something similar to the Stone-Cech compactification but in the category of compact metric spaces ?

 A: Neat question!  I haven't seen this construction before, but here are some thoughts.
First, I think separability of $X$ (and subsequently metrizability of $\mathcal{X}$) is really a red herring here.  For any bounded metric space $(X,d)$, you could construct a compactification in the same way, taking the closure $\mathcal{X}$ of the image of the map $X\to [0,N]^X$ that sends $x$ to the function $d(x,-)$ (where $N$ is some upper bound on the metric).  Equivalently, you could just take closure of the image of $X\to [0,N]^Q$ where $Q$ is any dense subset of $X$.  In the case that $X$ is separable, you can choose $Q$ to be countable, and so $\mathcal{X}$ will be metrizable, but this is not essential to what is going on here.
Here are some other ways to describe this compactification $\mathcal{X}$.  It is the smallest compactification of $X$ to which the function $d(x,-)$ extends continuously for each $x\in X$.  Alternatively, it is the spectrum of the closed subalgebra of $C_b(X)$ generated by the functions $d(x,-)$ for each $x\in X$ (these functions form a subspace of $C_b(X)$ which is canonically isometric to $X$).
Note that this compactification is highly sensitive to the choice of metric $d$, and not just how $d$ behaves at small scales.  For instance, consider the discrete space $X=\mathbb{N}$ with a metric $d$ such that $d(x,y)$ is always either $1$ or $2$ if $x\neq y$.  The points of $\mathcal{X}\setminus X$ correspond to all functions $\mathbb{N}\to\{1,2\}$ that are pointwise accumulation points of the functions $d(x,-)$.  So, by choosing $d$ appropriately, you can arrange for $\mathcal{X}\setminus X$ to be any nonempty closed subspaces of $\{1,2\}^\mathbb{N}$.  (This takes a bit of work to prove, since you have to be careful about the restriction that $d$ is symmetric.  Basically, you can choose $d$ freely "below the diagonal" of $\mathbb{N}\times\mathbb{N}$, and that is enough to control the accumulation points of the functions $d(x,-)$.)
