Every uniformity has a base of symmetric entourages

Choose $R\in\mathcal U$ with $R\circ R\subseteq E$, and put $S:=R\cap R^{-1}$.

Put $E_0:=E$. By finitely iterating the square-root axiom, choose entourages $E_1,\ldots,E_n$ such that $E_{k+1}\circ E_{k+1}\subseteq E_k$ for $0\le k<n$, and put $D:=E_n\cap E_n^{-1}$.

The set $S$ is an entourage, since $R,R^{-1}\in\mathcal U$ and a filter is closed under intersections; also $S=S^{-1}$ and $S\subseteq R\circ R\subseteq E$, because every entourage contains the diagonal.

The entourage $D$ is symmetric and $D\subseteq E_n$. Induction on $k$ gives $D^{\circ 2^k}\subseteq E_{n-k}$ for $0\le k\le n$, hence $D^{\circ 2^n}\subseteq E$. Since every entourage contains the diagonal and $n\le 2^n$, one may insert diagonal factors to obtain $D^{\circ n}\subseteq D^{\circ 2^n}\subseteq E$.

Thus symmetric entourages refine every entourage; their intersections are symmetric entourages and none is empty because each contains the diagonal, so they form a filter base by [L1].

Therefore symmetric entourages form a base and admit the asserted finite-composite control.