Let $x\in X$ and $S\subset X$ be countable and dense. Is there a sequence in $S$ converging to $x$?

Question

Let $(X,\tau)$ be a separable topological space and $S\subset X$ be a countable dense subset.

$(1)$ For any $x\in X$ can we find a sequence of elements in $S$ that converges to $x$?

$(2)$ For any non-isolated $x\in X$, can we find a sequence of elements in $X\setminus \{x\}$ that converges to $x$?

If $(X,\tau)$ is $T_1$ then $(1)\Rightarrow (2)$.

If $(X,\tau)$ is first countable then $(1)$ and $(2)$ hold.

If $\tau$ is the cofinite topology the $(1)$ and $(2)$ hold.

To prove $(1)$ if suffices to show the following:

Let $\mathcal{F}$ be a family of subsets of $\mathbb{N}$ that has the finite intersection property, there exists a sequence $(a_n)_n$ in $\mathbb{N}$ such that, for every $F\in\mathcal{F}$, sequence $(a_n)_n$ belongs eventually in $F$.

Answer 1

The statement you wish to prove is not true. A counterexample is given by the Stone-Čech compactification of $\mathbb N$, denoted $\beta \mathbb N$.

Since it is a compactification of $\mathbb N$, it follows that $\mathbb N$ itself is a (countable) dense subset.

One property of this space is that all convergent sequences are eventually constant, meaning that if $( x_n )_n$ is a sequence in $\beta \mathbb N$ which converges to $x \in \beta \mathbb N$, then there is an $N$ such that $x_n = x$ for all $n \geq N$. In particular, no point in the remainder $\beta \mathbb N \setminus \mathbb N$ is the limit of a sequence in $\mathbb N$.

More details about these facts can be found on Dan Ma's Topology Blog: Stone-Cech Compactification of the Integers – Basic Facts