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LetThis is the dense countable family of cofinite infinite subsets. Every ordinary basic open set $[s]$ contains the cofinite set $s\cup \{n:n>\max s\}$. Thus $D$ is $\tau$-dense: its ordinary closure is the whole space, so it is not $\tau$-nowhere dense.

It is $\ast$-closed. Indeed, for an infinite noncofinite $X$, the Ellentuck topology neighbourhood $[\varnothing ,X]$ contains only noncofinite sets, because every subset of $X$ misses the infinite complement of $X$. This neighbourhood is disjoint from $D$.

Finally every nonempty basic $\ast$-open set $[s,A]$ contains a smaller basic $\ast$-open set disjoint from $D$: enumerate $A$ and let $B$ consist of every second point. Then $A\setminus B$ is infinite, and $[s,B]\subseteq [s,A]$ contains no cofinite set. In particular $D$ has empty $\ast$-interior. As it is already $\ast$-closed, its $\ast$-closure has empty $\ast$-interior, proving that it is $\ast$-nowhere dense.

Take $\tau$ to be the ordinary topology on infinite subsets, whose basic cylinders $[s]$ fix a finite initial segment. LetThis is a dense countable family of cofinite infinite subsets. It is countable because finite complements form a countable family. Each singleton is $\tau$-closed: distinct increasing enumerations differ at some finite coordinate, which separates them by cylinders. No singleton has interior, since every cylinder admits more than one infinite continuation. Thus every singleton is $\tau$-nowhere dense, and $D$ is $\tau$-meagre.

On the other hand, every cylinder $[s]$ contains the cofinite set $s\cup \{n:n>\max s\}$, with $\max \varnothing =0$. Hence $D$ is $\tau$-dense, its closure is all of $X$, and

The paper does not define its symbol $\tau$. If the course instead uses it for the Ramsey cone topology, with basic neighborhoods $[A{]}^{\omega }$ and no finite stem, use $X$ itself as the example. For $D_j=\{Z:\min Z=j\}$, its cone-closure is $\{Z:j\in Z\}$: every cone neighborhood of a set containing $j$ contains an infinite subset of minimum $j$, while a cone based on a set omitting $j$ misses $D_j$. This closure has empty interior, because every infinite reservoir can be thinned to omit $j$. Thus every $D_j$ is cone-nowhere dense, and $X={\bigcup }_{j\ge 1}{D}_j$ is cone-meagre but is not cone-nowhere dense. This supplies the requested example under either convention, without conflating the two topologies.