My too-long-for-a-comment 2¢. It is not possible to build a theory of completion (be it of metric or uniform spaces) without using (uncountable $G_\delta$-sets in) $\mathbb R$ as a model. Results throughout are taken from A. S. Kechris, Classical Descriptive Set Theory, 1995.
Let us restrict our attention to separable metric (rather: metrizable, see below) spaces: if we cannot understand completion of these spaces without $\mathbb R$, then a fortiori we cannot do so for more general spaces (e.g. non-separable uniformizable).
It is not so relevant (to a negative answer to) the question, whether we can realize completions explicitly, or rather whether we can consider the question of the existence of complete metrics metrizing a given topology. In other words, the question in the op is intimately related to
Question: Can we build a theory of Polish spaces without $\mathbb R$?
(Recall that a Polish space is a separable metrizable topological space $(X,\tau)$ so that there exists a distance $d$ metrizing $\tau$ and additionally so that $(X,d)$ is complete as a metric space). The connection to the question in the op is a consequence of 1.: indeed, if $(X,d)$ is separable, then its completion $\overline X$ is complete, thus, in particular, Polish.
The advantage of using Polish spaces comes from the fact that they are at the foundation of descriptive set theory (but I will try to stay away from it as much as possible).
Once we have recast the problem as in 2., then we start seeing why $\mathbb Q$ is not a good model.
Fact 1: $\mathbb Q$ is not a Polish space.
That is, there exists no distance on $\mathbb Q$ (metrizing its standard topology as a subset of the real line and) turning it into a complete metric space. This is a topological fact, but topological facts on 'small' spaces have usually some connection with problems of cardinality. Indeed, we also have
Fact 2: Every countable metrizable space with no isolated points is homeomorphic to $\mathbb Q$.
Combining Facts 1 and 2 we see that a theory of infinite (in cardinality) Polish spaces requires uncountable sets in a very essential way.
- Recall that a $G_\delta$-set is any up-to-countable intersection of open sets. $G_\delta$-sets are relevant to the theory of Polish spaces because of the following
Theorem (Alexandrov): A subspace of a Polish space is itself Polish if and only if it is a $G_\delta$-set.
- Once $\mathbb Q$ is out of the way, let us turn to why $\mathbb R$ is necessary and sufficient.
Fact 3: $\mathbb R^\mathbb N$, the Hilbert cube $\mathcal H:=[0,1]^\mathbb N$, the Baire space $\mathcal N=\mathbb N^\mathbb N$, and the Cantor space $\{0,1\}^\mathbb N$ are Polish spaces. They are the prototypical Polish spaces in the sense that they satisfy some universal property along the following lines:
- (Kechris 4.17) Every Polish space is homeomorphic to a closed subset of $\mathbb R^\mathbb N$;
- (Kechris 4.14) Every Polish space is homeomorphic to a $G_\delta$-set in $\mathcal H$;
- I am not listing the properties of the Baire space and of the Cantor space since they are not relevant in the discussion below.
Fact 3.1. is the relevant one here: since every closed subset of a complete space is complete, every Polish space is homeomorphic to a complete metric subspace of $\mathbb R^\mathbb N$, i.e. one that is complete with a particular assigned metric.
A proof of both Facts 3.1 and 3.2 relies on the completeness of $\mathcal H$ and, implicitly, on that of the unit interval, thus again on the completeness of $\mathbb R$.
- Let us turn back to the problem of completion. Metric (and uniform) completions are characterized by the following universal property:
Fact 4 (completion): For every metric space $(X,d)$ there exists a metric space $(\overline X,\overline d)$ (unique up to isomorphism) and a dense embedding $\iota\colon X\to\overline X$ with the following property. For every complete metric space $(Y,\rho)$ and every uniformly continuous map $f\colon X\to Y$ there exists a unique (uniformly) continuous extension $\overline f\colon \overline X\to Y$ so that $f=\overline f \circ \iota$.
While seemingly unrelated to Polishness or $G_\delta$ sets, Fact 4 may be understood as the analogon in $\mathbf{Met}$ of the following result in $\mathbf{Top}$:
Theorem (Kuratowski, Kechris 3.8) Let X be metrizable and Y completely metrizable. Further let $A\subset X$ and $f\colon A\to Y$ be continuous. Then there is a $G_\delta$ set $G$ with $A\subset G\subset \overline A$ and a continuous extension $f_G$ of $f$ to $G$.
In turn, Kuratowski's Theorem is related to the realization of Polish spaces as subspaces of $\mathcal H$ by the following result of Lavrentiev.
Theorem (Lavrentiev, Kechris 3.9) Let $X,Y$ be completely metrizable, $A\subset X$, $B\subset Y$, and $f\colon A\to B$ be a homeomorphims. Then there exist $G_\delta$-sets $G\supset A$ and $H\supset B$ and a homeomorphism $h\colon G\to H$ extending $f$.
Note. While the above line of reasoning is slightly more implicit than necessary, this is essentially due to the additional of requirement of separability, which makes things more understandable but somewhat less polished.
If we drop separability and metrizability and we only understand completeness in the correct category $\mathbf{USp}$ of uniform spaces with uniform maps, then we have the following result of Shirota, which streamlines the above discussion:
Theorem (Shirota, Osaka J. Math., 1952) For a completely regular Hausdorff topological space, the following are equivalent:
- the natural uniformity of $X$ is complete;
- X is realcompact (equivalently: homeomorphic to a closed subset of a Cartesian product of real lines).
(See also A.6.10 in Borchers–Sen, Mathematical Implications of Einstein-Weyl Causality, 2006)
Again, completeness is characterized by the property of being a closed subspace of a product of real lines, as in Fact 3.1.
