Let $(X, \mathcal{U})$ be a compact Hausdorff uniform space. It is known that $\bigcap \{U: U\in \mathcal{U}\}= \Delta_X$.
Is there a sequence $\{U_n\}_{n\in\mathbb{N}}$ with $U_{n+1}\subseteq U_n$ and $\bigcap_{n\in\mathbb{N}}U_n=\Delta_X$?
Let $(X, \mathcal{U})$ be a compact Hausdorff uniform space. It is known that $\bigcap \{U: U\in \mathcal{U}\}= \Delta_X$.
Is there a sequence $\{U_n\}_{n\in\mathbb{N}}$ with $U_{n+1}\subseteq U_n$ and $\bigcap_{n\in\mathbb{N}}U_n=\Delta_X$?
No, this only happens if $X$ is metrisable. This follows from this answer, as then the diagonal is a closed $G_\delta$ in a normal space and hence a zero-set.
So take any compact Hausdorff non-metrisable $X$ for a counterexample e.g. Double Arrow ( $X \times \{0,1\}$ in the order topology induced by the lexicographic order) or $\omega_1+1$, or the one-point compactification of an uncountable discrete space, etc.