I'm studying point set topology using Munkres' Topology (2nd edition), and at some point he affirms that (Thm 30.1)
Let $X$ be a topological space. Let $A$ be a subset of $X$. If there is a sequence of points in $A$ converging to $x$, then $x \in \overline{A}$; the converse holds if $X$ is first-countable.
I would like to know if it's true that
If $x \in \overline{A} \subset X$ implies that there is a sequence in $A$ converging to $x$ for all $A$, then $X$ first-countable.
I cannot prove it nor find a counterexample.
It turns out that the answer is no.
Consider the space $X$ to be an uncountably infinite set with the co-finite topology. That is, the topology where the closed sets are exactly all the finite subsets of $X$ together with the whole of $X$.
X satisfies the sequence existence property
if $A\subseteq X$ is finite, than $\bar A = A$ so for every $x\in \bar A$ the constant sequence $(x)_{n\in \mathbb{N}}$ in $A$ converges to $x$.
if $A$ is infinite then we will take some countably infinite subset of $A$ and construct a sequence by enumerating its elements. This is a sequence in $A$ where every element is distinct, and such a sequence converges to every point $x\in X$!
X is not first countable
Suppose there is some countable basis $\{B_n\}_{n \in \mathbb{N}}$ for a point $x \in X$. For every $n$, the complement of $B_n$, which I will mark $C_n$ , is finite. And so their union: $$C:=\bigcup_{n\in \mathbb{N}}C_n$$ is countable as a countable union of finite sets. But $X$ is uncountable, so there must exist a point $p\in X$ different from $x$ such that $p \notin C$, which is equivalent to $p \in \bigcap_{n\in \mathbb{N}}B_n$.
Now consider the set: $$ U = X\setminus\{p\}$$ Clearly $x\in U$ and $U$ is open, and so $U$ is indeed an open neighborhood of $x$. But for every $n\in \mathbb{N}$ we have that $p\in B_n$ while $p \notin U$, and therefore $B_n \not\subseteq U$.
But that means $\{B_n\}_{n \in \mathbb{N}}$ cannot be a neighborhood basis of $x$ and we are done.