Let $(X_n,\pi_{n,k})$ be a countable projective system of separable topological spaces, where each $\pi_{n,m}$ is surjective. Suppose that $D_n$ is a dense subset of $X_n$.
Is there a reasonable/known description of the finest topology on the underlying set of $\varprojlim X_n$ making $D\triangleq \bigcap_{n} \pi_n(D_n)\neq \emptyset$ dense? Here $\pi_n$ is the canonical (surjective) map taking $X_n$ onto $\varprojlim X_n$.
Thoughts: Such a topology must exist, since the intersections of topologies is a topology and since the trivial topology on the underlying set of $\varprojlim X_n$ ensures that there is a non-empty set of topologies on the underlying set of $\varprojlim X_n$ satisfying the question.
Edit: It was brought to my attention that the original formulation of this question was already answered some time-ago in this post.
Note: This is the ~Dual of this post.
Essentially you ask the following question:
Given a non-empty subset $D$ of a set $X$, is there a finest topology $\tau$ on $X$ such that $D$ becomes dense in $(X,\tau)$?
The case $D = X$ is trivial, the discrete topology will do. Thus let us assume $D \subsetneqq X$.
If $D =\{x\}$ is a singleton, then we take $\tau$ to be the set of all $U \subset X$ such that $x \in U$ (plus the empty set). This has the desired property.
If $D$ has more than one point, then the answer is "no".
Let $x \in D$. Then $\tau_1 = \{\emptyset, X \setminus \{x\}, X \}$ and $\tau_2 = \{\emptyset, (X \setminus D) \cup \{x\}, X \}$ are two topologies such that $D$ is dense in $(X,\tau_i)$. Now assume that there is a finest topology $\tau$ with this property. Then $\tau_i \subset \tau$, thus $X \setminus \{x\},(X \setminus D) \cup \{x\}$ belong to $\tau$ and so does their intersection $U = (X \setminus \{x\}) \cap ((X \setminus D) \cup \{x\}) = X \setminus D$. But $U$ does not contain any point of $D$, thus $D$ is not dense in $(X,\tau)$, a contradiction.