a compact uniform space which is not metrizable

Let $(X, \mathcal{U})$ be a compact uniform space which is not metrizable and and $\{U_i\}_{i=0}^{\infty}$ be a countable set of $U_i\in \mathcal{U}$ with $U_{i-1}\subseteq U_i$ and $U_1\subseteq V$ that $V\in \mathcal{U}$ and $V\circ V\circ V\subseteq U$ for $U\in \mathcal U$. In Proof of Theorem 1 in https://projecteuclid.org/download/pdf_1/euclid.pjm/1103038056, author claimed that since $\{U_i\}_{i=1}^{\infty}$ is not a base for the uniformity $\mathcal{U}$, there is $W\in \mathcal{U}$ with $W\subseteq U$ such that for every $i>0$, $U_i\cap comp W\neq \emptyset$,

Q. Proof of $U_i\cap comp W\neq \emptyset$ is not clear for me, ( The notation $camp W$ is not define in this paper and it is not clear for me)

Thanks

$\operatorname{comp}W$ is just the complement of $W$ in $X$, which I'll denote by $W^c$ instead.

Recall that $\{U_i\}_{i=1}^\infty$ being a base (for entourages inside $U$) means:

$$\forall W \in \mathcal{U}( W \subset U) \implies (\exists i: U_i \subseteq W)\text{.}$$

So not being a base means, applying standard logic:

$$\exists W \in \mathcal{U}: (W \subset U)\land (\forall i: U_i \nsubseteq W)$$

and the last negation of inclusion can be written as $U_i \cap W^c \neq \emptyset$ (not all members of $U_i$ are in $W$, so some of them are in $W^c$...). So indeed:

$$\exists W \in \mathcal{U}: (W \subset U)\land (\forall i: U_i \cap W^c \neq \emptyset)$$

So the negation of being a base inside $U$ gives us the promised $W$.

(If the $(U_i)$ would be a base for entourages inside $U$, then for any $V \in \mathcal{U}$, $V \cap U \in \mathcal{U}$ would contain some $U_i$ so $V$ would contain a $U_i$ as well, and $\mathcal{U}$ would have a countable base and thus be metrisable, which it is not. So the negation holds.)