Let $\mathcal{U}$ be a uniform structure on a set $X$ We define a subset $A$ to be open if and only if for each $x \in A$ there exists an entourage $U \in \mathcal{U}$ such that whenever $(x,y) \in U$ then $y \in A$

Let us verify that this defines a topology on $X$

Clearly, the subsets $\emptyset$ and $X$ are open. If $A$ and $B$ are two open sets, then for each $x \in A \cap B$ there exist an entourage $U$ such that, whenever $(x,y) \in U$ then $y \in A$ and an entourage $V$ such that, whenever $(x,y) \in V$ then $y \in B$ Consider the entourage $U \cap V$ whenever $(x,y) \in U\cap V$ then $y \in A \cap B$ hence $A \cap B$ is open.

Suppose $\mathcal{F}$ is an arbitrary family of open subsets. For each $x \in \bigcup \mathcal{F}$ there exists $A \in \mathcal{F}$ such that $x \in A$ Let $U$ be the entourage whose existence is granted by the definition of open set. We have that whenever $(x,y) \in U$ then $y \in A$ hence $y \in \bigcup \mathcal{F}$ which concludes the proof.