Question

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 teh 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 $U$.