Let $X$ be an arbitrary non-empty set. Let $d$ be an arbitrary (non-trivial) metric on $X$. Let $\mathscr{T}$ be an arbitrary (non-trivial) topology on $X$.
Is it possible that $(X,d)$ and $(X,\mathscr{T})$ have different collections of dense subsets (under metric $d$ and under topology $\mathscr{T})$? In other words, is it possible that there exists a subset $S$ of $X$ such that $S$ is either dense in $(X,d)$ or $(X,\mathscr{T})$ but NOT both?
On
Yes. Let $X=\mathbb R$ and let $\mathscr T$ be the usual topology on $\mathscr R$.
Define a bijection $f:\mathbb R\to\mathbb R$ as follows:
$$f(x)=\begin{cases}
\ \ \ x\ \text{ if }\ x\in\mathbb Q,\\
-x\ \text{ if }\ x\notin\mathbb Q.
\end{cases}$$
Define the metric $d(x,y)=|f(x)-f(y)|$.
The set $S=(0,\infty)$ of all positive real numbers is dense in $(X,d)$ but not in $(X,\mathscr T)$.
Yes. Let $X = \{0,1\} \cup (2,3)$ with the metric inherited from the usual metric for the reals. The metric for $X$ is not trivial.
Give $X$ the topology with the open sets of all the open sets of $(2,3)$ as a subspace of the reals and the open set $\{0,1\}$. The topology is not trivial.
$D=\{0\} \cup (2,3)$ is a dense subset of the topological space but not of the metric space.