The sets containing an entourage ball about each of their points form a topology

The sets $\varnothing$ and $X$ are open: the first has no points to test, and for $x\in X$ every entourage ball is contained in $X$.

An arbitrary union of open sets is open, because a point in the union lies in one member and retains that member's entourage ball.

If $x\in O\cap P$, choose entourage balls $E[x]\subseteq O$ and $F[x]\subseteq P$; then $(E\cap F)[x]\subseteq O\cap P$, so binary intersections are open.

By steps 1.1 to 1.3, the open sets form a topology by [L1].

Let $E$ be an entourage and define $O_E=\{y\in E[x]:F[y]\subseteq E[x]\text{ for some }F\in\mathcal U\}.$ This set is open. Indeed, given $y\in O_E$, choose $F$ as displayed and then a symmetric $G$ with $G\circ G\subseteq F$. If $z\in G[y]$, symmetry gives $G[z]\subseteq(G\circ G)[y]\subseteq F[y]\subseteq E[x]$, so $z\in O_E$; hence $G[y]\subseteq O_E$. Now choose a symmetric $D$ with $D\circ D\subseteq E$. If $y\in D[x]$, then $D[y]\subseteq E[x]$, so $y\in O_E$. Thus $x\in D[x]\subseteq O_E\subseteq E[x]$, proving that $E[x]$ is a neighbourhood of $x$.

Conversely, if $N$ is a neighbourhood of $x$, it contains an open set $O$ with $x\in O$; the definition of the topology supplies an entourage $E$ with $E[x]\subseteq O\subseteq N$. Thus the entourage balls refine every neighbourhood, and by step 3.1 they are themselves neighbourhoods. They form a neighbourhood base by [L2].