First, a small motivation: Suppose we are looking for a compactification of uniform spaces, satisfying an universal property similar to the one of the Stone-Čech compactification of a locally compact space, which may be regarded as the spectrum of the C*-algebra of continuous bounded complex-valued functions on the space.
Now, let $X$ be a separated uniform space (hence Hausdorff as a topological space). If we wanted to construct a compactification of $X$ is a similar manner as the one described above for $\beta X$, we'd proceed as follows: Let $UC^b(X)$ denote the space of complex-valued, bounded, uniformly continuous function on $X$, and let $\sigma X$ be its spectrum. Then $X$ embeds naturally in $\sigma X$ via the map $i:X\mapsto\sigma X$, $i(x)(f)=f(x)$.
Uniform spaces satisfy a kind of "uniform Tychonoff property": If $U$ is a neighbourhood of $x$, then there exists a uniformly continuous function $f$ such that $f(x)=1$ and $f=0$ outside $U$ (this is easily seen by considering uniformly continuous pseudometrics which generate the uniform structure). This implies that the map $i:X\to\sigma X$ is in fact a topological embedding. Since $UC^b(X)$ is unital, $\sigma X$ is compact. Hence, for $\sigma X$ to be a compactification of $X$, it remains only to obtain a positive answer for the following question:
Question: Is $X$ dense in $\sigma X$?
I tried following the proof from Murphy's book that, for compact $K$, the spectrum of $C(K)$ is $K$ itself, but it does not translate nicely to this case (apparently).
Any help is appreciated.
Take a look at Definition 6.13 and the development thereafter in Warren Page's Topological Uniform Structures. He explains there that, if a uniform space is to have a uniform compactification in which it is dense, then the uniform space must be totally-bounded. In this case, the compactification and completion coincide. In particular, it seems that the reason you are having difficultly proving density is that it is not always true---it seems that it will be true iff $X$ is totally-bounded.