A normal sequence of entourages yields a uniformly continuous pseudometric with controlled dyadic balls

For $x,y\in X$, let $W(x,y)$ be the set of sums $\sum_{i<k}2^{-n_i}$ over all finite chains $x=x_0,\ldots,x_k=y$ with $(x_{i-1},x_i)\in E_{n_i}$. This set is nonempty because the one-edge $E_0=X\times X$ chain has weight $1$. Every dyadic term is positive by [L4], so every such finite sum is nonnegative by [L3]; hence $0$ is a lower bound. By [L2], the infimum exists; define $p(x,y):=\inf W(x,y)$.

We prove by strong induction on the number $k$ of edges, simultaneously for every $n$, that a $k$-edge chain of total weight less than $2^{-n}$ has $E_n$-related endpoints. For $k=0$ the endpoints coincide and hence are $E_n$-related by [A2]. Now let $k\ge1$ and assume the claim for every shorter chain. Its total weight $w$ is positive by [L3] and [L4]. Take the first edge for which the cumulative weight through that edge exceeds $w/2$. The subchain before it has weight at most $w/2$, and the subchain after it has weight less than $w/2$; because $w<2^{-n}$, both are less than $2^{-(n+1)}$. They have fewer than $k$ edges, so the induction hypothesis makes both endpoint pairs $E_{n+1}$-related. The middle edge has weight $2^{-m}\le w<2^{-n}$; strict decrease of the dyadic weights gives $m\ge n+1$, and decreasingness of $(E_j)$ puts that edge in $E_{n+1}$. Thus the endpoints lie in $E_{n+1}^{\circ3}\subseteq E_n$. Strong induction proves the claim for every finite chain.

The empty chain has weight $0$, while all weights are nonnegative, so $p(x,x)=0$. Reversing a chain preserves its weight because each $E_n$ is symmetric, so $p(x,y)=p(y,x)$. For the triangle inequality, suppose instead that $p(x,z)>p(x,y)+p(y,z)$ and put $\delta=(p(x,z)-p(x,y)-p(y,z))/3>0$. By [L2], choose an $x$-to-$y$ chain of weight $a<p(x,y)+\delta$ and a $y$-to-$z$ chain of weight $b<p(y,z)+\delta$. Their concatenation has weight $a+b<p(x,y)+p(y,z)+2\delta<p(x,z)$ by [L3], contradicting that $p(x,z)$ is a lower bound of $W(x,z)$. Thus the triangle inequality holds, and $p$ is a pseudometric by [L1].

A one-edge $E_n$-chain has weight $2^{-n}$, so $E_n\subseteq\{p\le2^{-n}\}$.

If $p(x,y)\le2^{-n}$ with $n\ge1$, then [L4] gives $p(x,y)\le2^{-n}<2^{-(n-1)}$. Apply the epsilon property in [L2] with $\varepsilon=2^{-(n-1)}-p(x,y)>0$ to obtain a chain of weight less than $2^{-(n-1)}$. Step 1.2 gives $(x,y)\in E_{n-1}$. Thus $\{p\le2^{-n}\}\subseteq E_{n-1}$.

Given $\varepsilon>0$, choose $n$ with $2^{-n}<\varepsilon$ by [L4]. Then $E_n\subseteq\{p\le2^{-n}\}\subseteq\{p<\varepsilon\}$, so the latter set is an entourage by upward closure. By A gauge of pseudometrics and, on a nonempty set, the uniformity it generates, $p$ is uniformly continuous for the original uniformity.