Counter-example: a topology that is not first countable where elements in the closure are exactly the elements that are limits of sequences? Question 10A in Willard's General Topology is the following:

For each of the following spaces, answer these questions: Which sequences converge to which points? Is $X$ first countable? Does the result of Theorem 10.4 hold true for $X$? (One of your answers should show that first countability is not necessary in Theorem 10.4)

The statement of Theorem 10.4 is the following:

If $X$ is a first countable space and $E \subseteq X$, then $x \in
 \overline{E}$ if and only if there is a sequence $(x_n)$ contained in $E$ which converges to $x$.

The spaces we are asked to consider are the following:

*

*$X$ any uncountable set with the cofinite topology.

*$X$ any uncountable set with the cocountable topology.

*$X$ the real line with the topology in which the open sets are the sets of the form $(a, \infty)$, $a \in \mathbb{R}$.

*$X$ the Sorgenfrey line $\mathbb{E}$.

*$X$ any discrete space.

*$X$ any trivial space.

I am a bit stuck with trying to explain if the result of Theorem 10.4 holds for one of the spaces that is not first countable. If I am correct, each of the topological spaces in 3-6 are all first countable, so any these cannot be the space that shows that first countability is not necessary in Theorem 10.4, so it has to be $X$ an uncountable set with either the cofinite or cocountable topology.
If $X$ has the cofinite topology, the convergent sequences are those for which (1) there exists no value which the sequence takes infinitely many times (and these sequences converge to any value $x \in X$) or (2) there exists exactly one value which the sequence takes infinitely many times (and the sequence converges to this infinitely repeating value).
If $X$ has the co-countable topology, the convergent sequences are those which are eventually constant (and the sequence converges to this constant value).
I have tried playing around with various definitions but I'm having trouble seeing which of these two satisfy, "if $E \subseteq X$, then $x \in \overline{E}$ if and only if there is a sequence $(x_n)$ contained in $E$ which converges to $x$."
Am I missing another possible option (is one of the other spaces listed not first countable)? Or is there a straightforward argument that shows this theorem holds for $X$ uncountable with either the cofinite or cocountable topology?
Thanks in advance!
 A: $\newcommand{\cl}{\operatorname{cl}}$You are correct about which of the spaces are first countable. For the other two you need to ask yourself not only which sequences converge, but also what the closed sets are. Let $X$ be uncountable. If $X$ has the cofinite topology, the only closed sets are $X$ itself and its finite subsets, so for any $E\subseteq X$ we know that $\cl E=E$ if $E$ is finite, and $\cl E=X$ if $E$ is infinite.

*

*If $E$ is finite, the only convergent sequences in $E$ are the ones that are eventually constant; they converge to points of $E$, and every point of $E$ is the limit of such a sequence, so $x\in\cl E$ iff some sequence in $E$ converges to $x$.

*If $E$ is infinite, it has a sequence of distinct points, and that sequence converges to every $x\in X=\cl E$, so again $x\in\cl E$ iff some sequence in $E$ converges to $x$.

Spaces with this property are Fréchet-Urysohn spaces; Theorem 10.4 says that all first countable spaces are Fréchet-Urysohn.
Now suppose that $X$ has the co-countable topology, and let $E\subseteq X$. The only convergent sequences in $E$ are the ones that are eventually constant, and they converge to points of $E$. If $E$ is countable, this is fine: in that case $\cl E=E$, and a point is in $\cl E$ iff some sequence in $E$ converges to it. If $E$ is an uncountable proper subset of $X$, however, $\cl E=X$, but it’s still true that the only limits of sequences in $E$ are the points of $E$. This space is therefore not Fréchet-Urysohn.
