Q:

What is a first countable, limit compact space that is not sequentially compact?

I just read a proof that adds the assumption of T1 to conclude that a first countable, limit compact and T1 space must be sequentially compact, but I didn't understand what happens if we drop the T1 assumption.
If $X$ is limit point compact and first countable, and $\{a_n\}$ is a sequence in $X$, then it has a limit point $x$.
By first countability, we can find a basis of neighborhoods of $x$ and then build a subsequence that converges to $x$. Isn't that enough?
Reference: The proof is from relationship among different kinds of compactness by rm50 (theorem 2). As you can see, $q$ appears from nowhere.

A:

Consider the right topology on $\mathbb{R}$: the topology generated by the sets of the form $$(a,\infty) = \{ x \in \mathbb{R} : x > a \}$$ for $a \in \mathbb{R}$.

It is clearly first- (even second-) countable. (Consider only rational $a$ for the basic open sets.)
If $ A\subseteq \mathbb {R} $ is nonempty, then any $x \in \mathbb{R}$ which is (strictly) less than an element of $A$ is a limit point of $A$, hence it is limit point compact.
The sequence $\langle -n \rangle_{n \in \mathbb {N}}$ has no convergent subsequence, hence it is not sequentially compact.

Note that in T1-spaces, to be a limit point of a set every neighbourhood must contain infinitely many points of that set. This is what allows for the construction of the subsequence.

If $\langle x_n \rangle_{n \in \mathbb {N}}$ is a sequence in $X$, then either it has a constant subsequence (which obviously converges), or the set $A= \{ x_n : n \in \mathbb {N} \}$ is infinite, and hence has a limit point, $ x $.
Fix a decreasing countable base $\{ V_k \}_{k \in \mathbb{N}}$ for $x$.
Recursively pick $n_k \in \mathbb{N}$ ($k \in \mathbb{N}$) so that $x_{n_k} \in V_k$ and $x_{n_k} > x_{n_{k-1}}$. This can be done because $V_k \cap A$ is infinite, and thus so is the set $\{ n \in \mathbb {N} : x_n \in V_k\}$. (Or, to go a bit more slowly, since $X$ is T1 finite sets are closed, and so $V_k \setminus \{ x_n : n \leq n_{k-1}, x_n \neq x \}$ is an open neighbourhood of $x$, and so contains an element $z$ of $A$ distinct from $x$. Since $z\in A$ there must be an $n$ such that $z=x_n$, and by choice of neighbourhood of $x$ it must be that $n > n_{k-1}$, so set $n_k = n$.)
The subsequence $\langle x_{n_k} \rangle_{k \in \mathbb {N}}$ then converges to $x$.

However for non-T1-spaces a neighbourhood of a limit point may only contain a single point of the set, which makes the construction of a subsequence as in the third paragraph of the above demonstration problematic.
In the above space, the only open set which contains infinitely many points of the sequence $\langle -n \rangle_{n \in \mathbb {N}}$ is the entire space $\mathbb {R}$, although $\overline { \{ -n : n \in \mathbb {N}\}} = (-\infty,-1]$.

