Corollary

Let $A$ be a compact set in a uniform space $X$; then as $V$ runs through the set of entourages of $X$, the sets $V(A)$ form a fundamental system of neighbourhoods of $A$.

Let $U$ be any open neighbourhood of $A$, then $B = \complement U$ is closed and does not meet $A$; hence, by Proposition 4, there is an entourage $V$ such that $V(A) \cap V(B) = \varnothing$, and therefore $V(A) \subset U$.