For every uniformizable space $X$ there is a finest uniformity on $X$ compatible with the topology of $X$ called the fine uniformity or universal uniformity.

To construct Fine uniformities, Let the family $\{\mathcal{V}_i:i\in I\}$ of all compatible uniformities on $X$ and generates the initial uniformity $\mathcal U$ w.r.t. $id_i:X\to(X,\mathcal{V}_i)$ on $X$. therefore, $\mathcal U$ is finer than each $\mathcal{V}_i$ for $i\in I$. Let us note that, completely regularity is used to show that $\{\mathcal{V}_i:i\in I\}\not=\emptyset$, since $X$ is a completely regular(uniformizable) space and have a compatible uniformity.

Exercise 36H. from Willard, General Topology says that,

The fine uniformity $\mathcal D_F$ on a uniformizable space is the uniformity having for a base $\beta$, the open sets $D\supset\Delta$ such that for some sequence $D_1, D_2,...$ of open sets containing $\Delta$, $D_n\circ D_n\subset D_{n-1}$ for all $n$ and $D_1= D$.

Is $\mathcal U$ and $\mathcal D_F$ exactly the same uniformity on a uniformizable space $X$? If the answer is yes, then how we can construct a base $\beta$ for $\mathcal U$ ?

Edit: I am trying to solve this question. $\beta$ is a base for a uniformity on $X$ and the uniformity $\mathcal D_F$ generated by $\beta$ is finer than $\mathcal U$. now it is enough to show that, $\mathcal D_F\subset \mathcal U$. If the $\mathcal D_F$ is compatible with the topology of $X$, then $\mathcal D_F\subset \mathcal U$, since $\mathcal U$ is the finest uniformity on $X$ compatible with the topology of $X$.

Now my problem is to show that $\mathcal D_F$ is compatible with the topology of $X$. thanks in advice