A uniform space $(S,\mathcal{U})$ is called separated iff for every $x\neq y$ in $S$, there is a $U\in\mathcal{U}$ with $\langle x,y\rangle\notin U$. Show that if $S$ has more than one point, a separated uniformity never converges as a filter for its product topology.
My effort:
$\mathcal{U}$ is a filter in $S\times S$. Assume it converges to $\langle a,b\rangle$, i.e., every neighborhood of $\langle a,b\rangle$ is an element of $\mathcal{U}$. We are using product topology for $S\times S$. Thus the neighborhood of $\langle a,b\rangle$ is in the form of $N_a\times N_b$ where $N_a$ and $N_b$ are neighborhoods of $a$ and $b$, respectively, both in the topology $\mathcal{T}$ on $S$. $\mathcal{T}$ is just the uniform topology defined by $\mathcal{U}$, which is the collection of all sets $V\subset S$ such that for each $x\in V$, there is a $U\in\mathcal{U}$ such that $\{y:\langle x,y\rangle\in U\}\subset V$. We will use the following three properties of uniformity to derive a contradiction.
(1) $N_a\times N_b$ includes the diagonal $D:=\{\langle s,s\rangle:s\in S\}$. Specifically, both $\langle a,a\rangle$ and $\langle b,b\rangle$ are in $N_a\times N_b$. Thus $a\in N_b$ and $b\in N_a$.
(2) $(N_a\times N_b)^{-1}=N_b\times N_a\in\mathcal{U}$.
(3) There is a $B \in\mathcal{U}$ with $B\circ B\subset N_a\times N_b$.
Suppose $a\neq b$. Using the fact that the uniform topology of a separated uniformity is always Hausdorff, there exist $N_a$ and $N_b$ such that $N_a\cap N_b=\emptyset$. This violates (1). So $a=b$.
Then I don't know how to proceed.
You need to make use of the fact that $\mathscr{U}$ is separated.
Let $x$ and $y$ be distinct points of $S$. Since $\mathscr{U}$ is separated, there is a $U_x\in\mathscr{U}$ such that $\langle x,y\rangle\notin U_x$, and hence $\{z\in S:\langle x,z\rangle\in U_x\}$ is a nbhd of $x$ that does not contain $y$. Since $x$ and $y$ were arbitrary, $S$ is $T_1$. (With a little more work one can show that $X$ is Hausdorff, but we don’t need that.)
Now let $\langle a,b\rangle\in S\times S$. $S$ has more than one point, so let $p\in S\setminus\{a\}$. Then $a$ has an open nbhd $N_a$ that does not contain $p$. But then $N_a\times S$ is an open nbhd of $\langle a,b\rangle$ that does not contain $\langle p,p\rangle$, so $N_a\times S\nsupseteq D$, and therefore $N_a\times S\notin\mathscr{U}$, and $\mathscr{U}$ does not converge to $\langle a,b\rangle$. Since $\langle a,b\rangle$ was an arbitrary point of $S\times S$, this shows that $\mathscr{U}$ does not converge.