Let $(S,<)$ be a countably infinite semilattice such that:
1) $S$ is no-where dense - i.e. there does not exist a subset $T$ such that for all $a,b\in T$ with $a<b$, there exists $c\in T$ with $a<c<b$,
2) $S$ has finite width $n$ - i.e. the maximum size of the anti-chains of $S$ is $n$.
Let $a,b\in S$ be such that $a<b$.
Question: Is the set $\{x\in S :a<x<b \}$ finite?
I think the simplest example is given by the ordinal $\omega+1=\{0<1<2<\cdots<\omega\}$. As a well order it contains no antichains, and it is well founded (hence nowhere dense), but of course $\{x\in\omega+1\mid 0<x<\omega\}$ is infinite. Ever infinite well order would do. You will find plenty of examples in the larger class of well quasi-orders too.