Let $(X,d)$ be a metric space and let $(X^*,d^*)$ denote its completion (via equivalence classes of Cauchy sequences in $X$). Let $f:X \to X^*$ be an isometry such that $f(X)$ is dense in $X^*$.
Now define $$ X'=(X^*\setminus f(X))\cup X $$ and a map $g\colon X'\to X^*$ by $$ g(x)=\begin{cases} f(x) & x\in X \\ x & x\notin X \end{cases} $$ Finally define, for $x,y\in X'$, $d'(x,y):=d^*(g(x),g(y))$.
Then $X'$ is a completion of $X$ (under the inclusion map) and $X\subset X'$.
This works as long as $(X^*\setminus f(X))\cap X =\emptyset$ (to show surjectivity of $g$). Can I show that this is the case in ZFC? There is a related question here.
Background for this question: The same idea arises when constructing the completion of a dense unbounded linearly ordered set via Dedekind cuts. In this case the disjoint condition follows from the axiom of foundation:
If $x\in X=A\cup B$ and $x=(A,B)$ then either $x\in A \in \{A\}\in x$ or $x\in B \in \{A,B\}\in x$.
Yes, in fact $X^*\cap X=\emptyset$. Lets write down explicitly:
$$X^*=Cauchy(X)/\sim$$
where $Cauchy(X)$ denotes the set of all Cauchy sequences on $X$ and $\sim$ is some relation. In other words $X^*\subseteq \mathcal{P}(Cauchy(X))$ as sets, where $\mathcal{P}(A)$ denotes the power set of $A$. Now what is a sequence? A sequence $(x_n)$ is actually a function $x:\mathbb{N}\to X$. And what is a function? It is a special subset of $\mathbb{N}\times X$, i.e. $x\in \mathcal{P}(\mathbb{N}\times X)$. And therefore $X^*\subseteq\mathcal{P}(\mathcal{P}(\mathbb{N}\times X))$ as sets.
We need to dive deeper. What is Cartesian product? Or rather: an ordered pair. It can be defined in many different ways, e.g. $(a,b):=\{\{a\},\{a,b\}\}$ following Kuratowski. Regardless, $a\in\in (a,b)$ and $b\in\in (a,b)$ where "$\in\in$" denotes "chain of belongings", i.e. in the Kuratowski variant $a\in \{a\}\in (a,b)$ and $b\in \{a,b\}\in (a,b)$. As far as I know all definitions will have that true. Edit: thanks to @Izaak van Dongen comment I learned that this is false. So we will only consider "standard" definitions where this holds. Note that this is also a fragile point of your argument about Dedekind cuts.
So now assume $[x_n]\in X^*$ is a Cauchy sequence class and assume $y=[x_n]$ belongs to $X$. Then we can replace $x_0$ with $y$ to get an equivalent Cauchy sequence. What that means is that $$y\in\in (0,y)\in (x_n)\in [x_n]=y$$ which cannot happen, because otherwise we would get an infinite nesting of sets (due to a loop), which is prevented by the axiom of regularity of ZF.
Note that this is a very subtle argument, the core observation is that we can replace the initial element of any Cauchy sequence with any other and still get the same Cauchy class. And another is the definition of ordered pair: I think it is possible to introduce a custom definition for which the result will be different (which btw will also make your argument on Dedekind cuts invalid). Also if we would consider some other $\sim$ then it actually might happen that $X^*\cap X\neq\emptyset$, for similar reason why $X\cap\mathcal{P}(X)$ might be nonempty, e.g. when $X=\{\{0\},0\}$.
That being said, it is worth noting that often it is more convenient to talk about abstractions, not concrete models. Meaning we can define $X^*$ and ordered pairs axiomatically. In such case it is not possible to prove what you want, it can be modelled to be true or false. Therefore you should just use $X^*$ and identify $X$ as its subset via embedding. It is safer, model independent and easier to understand.