Assuming dependent choice, every entourage admits a normal symmetric sequence subordinate to it

Given a symmetric entourage $E$, choose a symmetric $R$ with $R^{\circ2}\subseteq E$, and then a symmetric $D$ with $D^{\circ2}\subseteq R$. Since every entourage contains the diagonal, $D\subseteq R$, and hence $D^{\circ3}\subseteq R^{\circ2}\subseteq E.$ Thus there exists a symmetric $D\subseteq E$ with $D^{\circ3}\subseteq E$.

The relation $E\mathrel R D$ meaning that $D$ is symmetric, $D\subseteq E$, and $D\circ D\circ D\subseteq E$ is serial by step 1.1.

Choose a symmetric entourage $E_1\subseteq U$ using [L1]. Dependent choice applied to the serial relation of step 2.1 starting at $E_1$ gives $E_1,E_2,\ldots$. Adjoin $E_0=X\times X$; then $E_1^{\circ3}\subseteq X\times X=E_0$, and all the required properties hold.