I'm looking for such pathological examples of uniform spaces which are not metrizable, but whose underlying topology is metrizable. Willard in his General Topology text constructs such a uniformity using ordinals. I am asking for examples which do not rely on ordinals.


EDIT: Below is an example by Daniel Tausk using families of pseudometrics. I forgot to mention that, if possible, I would like an example that uses the definition through a diagonal uniformity, not the definition through pseudometrics nor the definition through uniform covers. See [Cryptomorphisms](https://mathoverflow.net/questions/15731/cryptomorphisms) for some elaboration on this phenomenon.



Nonmetrizable uniformities with metrizable topologies
The family of **all** neighbourhoods of the diagonal of $\mathbb{R}$ (with its normal topology) is a uniformity without a countable basis but it generates the normal topology.
The normal topology of $\mathbb{R}$ is the topology generated by the usual distance.
To see that the open sets in the plane that contain the diagonal generate a uniformity let $U$ be such an open set. For each $x$ there is $r\_x$ such that $B(x,r\_x)^2\subseteq U$. Now let $V$ be the union of the squares $B(x,\frac13r\_x)^2$; this is again an open set and it is not hard to show that $V\circ V\subseteq U$. 
If $U$ is an open set in the plane that contains the diagonal then the vertical section $U[x]$ is open in $\mathbb{R}$, so this uniformity generates at best a subtopology of the normal topology and hence certainly not the discrete topology. To see that it generates the normal topology consider, given $x$ and $r$, the open set $U\_{x,r}=B(x,r)^2\cup(\mathbb{R}\setminus\lbrace x\rbrace)^2$; then $U\_{x,r}[x]=B(x,r)$.
The uniformity does not have a countable base: if $\langle U\_n:n\in\mathbb{N}\rangle$ is a sequence of neighbourhoods of the diagonal then you can use diagonalisation to produce a neighbourhood such that $U\_n\not\subseteq U$ for all $n$: make sure that the square $[n,n+1]^2$ contains a point in $U\_n$ but not in $U$.


