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.