Let $X$ be a separable topological space. By definition, it has a countable dense subset $S$. Suppose $S$ is infinite. My question is:
What are the examples of such $X$ satisfying the property that for any $n\in\mathbb{N}$ and any $s_1,\ldots,s_n\in S$, the set $S\setminus\{s_1,\ldots,s_n\}$ remains a dense subset of $X$?
To explain the motivation behind let me first describe one such example: Let $(X,\lVert\cdot\rVert)$ be a separable normed vector space (over $\mathbb{R}$ or $\mathbb{C}$), and let $S$ be a countably infinite dense subset. One can show that for any $s\in S$, the subset $S\setminus\{s\}$ is still dense in $X$. The key is to show that every open ball centered at $s$ must intersect $S\setminus\{s\}$. This can be shown by contradiction, with the idea clear from drawing pictures. Suppose $\exists\delta>0$ such that $B(s,\delta)$ is disjoint from $S\setminus\{s\}$. Pick arbitrarily a unit vector $v$, and consider the element $x:=s+(\delta/2)v$. Then the ball $B(x,\delta/2)$ is contained in $B(s,\delta)$ but does not contain $s$, thus it is disjoint from the whole $S$. This contradicts the assumption that $S$ is dense in $X$.
Consequently, given any $y\in X$ and any $\epsilon>0$, if it happens that $s\in B(y,\epsilon/2)$, then by the result above $B(s,\epsilon/2)$ must contain some $s'\in S\setminus\{s\}$, thus $s'\in B(y,\epsilon)$ by triangle inequality. This shows that $S\setminus\{s\}$ remains to be dense in $X$.
(Side note: As an immediate consequence, the countable dense subset can therefore always be chosen to contains only nonzero vectors. This plays a key step, e.g. in (one of the approaches) showing that any separable Banach space can be isometrically embedded as a linear subspace into $\ell^{\infty}$.)
By induction, it follows that for any $n\in\mathbb{N}$ and any $s_1,\ldots,s_n\in S$, the subset $S\setminus\{s_1,\ldots,s_n\}$ remains to be dense in $X$. This shows that $(X,\lVert\cdot\rVert)$ is such an example.
Now, since the notion of separability is a topological one, it seems natural to think about the generalization of the results above to general topological space. At first I was naively thinking that the argument above can be generalized directly, simply by replacing the open balls to general open neighbourhood (balls require the notion of metric, which is not present in general topological space, but open sets are fundamental notion). However, soon I found that the problem is not so trivial. For example, consider the discrete topological space $X$ (with countably infinite underlying set). Such $X$ is separable, but it is well known that the only dense subset of $X$ is itself, so clearly any proper subset of it cannot remain to be dense in $X$. Such $X$ provides a simple counterexample for the result to hold in general.
Taking a closer look we may find that one of the reason it fails on discrete space is that the topology is too fine/large. For example, in the case of normed vector space, within the small open ball $B(s,\delta)$ we can always find another strictly smaller open ball $B(x,\delta/2)$, but in the discrete space it may happen that the open neighbourhood of $s$ turns out to be the singleton $\{s\}$ itself. From this, it seems that one necessary condition is to require the topological space satisfies the property that:
Given any point $x\in X$, any open neighbourhood of $x$ must contain some element $y\neq x$.
Moreover, in the case of normed vector space, the contradiction is achieved by constructing the ball $B(x,\delta/2)$ which does not intersect $s$. Thus, even though we seek for a topology that is not so fine, it must also be not too coarse. For example, another necessary condition would seems to be that
$X$ is Hausdorff.
In this question I would like to see some example of such topological space where the generalization is valid. If anyone happens to know the most general class of topological spaces where such generalization holds, please let me know as well!
Any comment or answer is welcome and greatly appreciated.
It is sufficient that $X$ be $T_1$ and have no isolated points (i.e. every singleton is closed and not open). Then for any non-empty open $U$ and finite $F$ contained in $X$, $U \setminus F$ is non-empty and open and so meets $S$. But then $U$ must meet $S \setminus F$. Hence $S \setminus F$ is dense.