Proposition 4

In a uniform space $X$, let $A$ be a compact set and $B$ a closed set such that $A \cap B = \varnothing$. Then there is an entourage $V$ of $X$ such that $V(A)$ and $V(B)$ do not intersect.

If the proposition were false, then none of the sets $A \cap \overline{V}(B)$, where $V$ runs through the set of symmetric entourages of $X$, would be empty; hence these sets would form a filter base on $A$, which would have a cluster point $x_0 \in A$. Hence for each symmetric entourage $V$ of $X$, $\overline{V}(x_0)$ would meet $B$ and therefore, as $B$ is closed, we should have $x_0 \in B$, contrary to hypothesis.