Definition 5.1.10 (Topology of a Uniform Space, [Proposition 2.1.2, Bou13]). Let $(X, \fU)$ be a uniform space and
then there exists a unique topology $\topo \subset 2^{X}$ such that $\cn_{\topo} = \cn$, known as the topology induced by the uniform structure $\fU$.
Proof. Using Proposition 4.4.4, it is sufficient to show that $\cn(x)$ is non-empty for all $x \in X$, and that it satisfies (F1), (F2), (V1), and (V2). Firstly, since $\fU \ne \emptyset$, $\cn(x) \ne \emptyset$ for all $x \in X$.
(F1): Let $U \in \fU$ and $V \supset U(x)$, then
As $\fU$ satisfies (F1), $W \in \fU$. Thus
(F2): Let $U, V \in \fU$, then $U(x) \cap V(x) = (U \cap V)(x)$. As $\fU$ satisfies (F2), $U \cap V \in \fU$ and $(U \cap V)(x) \in \cn(x)$.
(V1): Let $U \in \fU$. By (U1), $\Delta \subset U$, so $x \in U(x)$.
(V2): Let $U \in \fU$ and $x \in X$. By (U2), there exists $W \in \fU$ such that $W \circ W \subset U$. Let $y \in W$, then for any $z \in W(y)$, $(x, z) \in V$. Hence $W(y) \subset V(x)$ for all $y \in W$, so $V(x) \in \cn(y)$ for all $y \in W$.$\square$