Long ago, I learned about the Cauchy completion of metric spaces via the usual explicit construction of quotienting the set of Cauchy sequences. For a metric space $X$, let $\hat X$ denote this Cauchy completion of $X$. We then have a canonical isometric embedding $\iota\colon X\to\hat X$ ($x\mapsto(x, x, x, \ldots)$). It's easily seen that $\iota(X)$ is dense in $\hat X$, and using this, one can show that the following universal property characterizes $\hat X$ up to bi-isometries:
Any isometric embedding of $X$ into a complete metric space factors uniquely through $\iota$.
Now, my question is this: Using this as the definition of a Cauchy completion, and avoiding the Cauchy-sequence-construction, can one show that $X$ is dense in $\hat X$?
Let $Y$ be the closure of $X$ in $\hat{X}$. Then $Y$ also has the universal property of $\hat{X}$: existence follows by restricting the extension to $\hat{X}$ and uniqueness follows since $X$ is dense in $Y$. It follows by the usual uniqueness argument for objects with a universal property that the inclusion map $Y\to\hat{X}$ is an isomorphism (in the category of metric spaes and isometric embeddings), i.e. $Y$ is all of $\hat{X}$.