Spinning things out from a recent question, let's recall that every second countable sequentially compact space is compact, because every Lindelof countably compact space is compact.
We cannot weaken second countable to first countable; the first uncountable ordinal is a direct example of a first countable sequentially compact space which is not compact.
We similarly cannot weaken second countable to weakly Lindelof (every open cover has a countable subcollection whose union is dense); the altered long ray is weakly Lindelof and sequentially compact, but not compact. The pi-Base lacks a proof for weakly Lindelof, so let me share one: in fact, I'll show every open cover has a finite subcollection whose union is dense.
Let $X=Y\cup\{\infty\}$ be the altered long ray, where $Y=\omega_1\times[0,1)$ is an open subspace with its lexicographic topology, and each neighborhood of $\infty$ contains some $U_\alpha=\{\infty\}\cup\Big(\omega_1\setminus\alpha\Big) \times (0,1)$.
Take an open cover $\mathcal U$, and choose $\alpha<\omega_1$ where $U_\alpha\subseteq V\in \mathcal U$. We note that $\alpha\times [0,1)\cup\{\langle \alpha,0\rangle\}\cong[0,1]$ is compact, so choose $\mathcal F\subseteq\mathcal U$ finite covering it. Then $\mathcal F\cup\{V\}$ covers $X\setminus(\omega_1 \times \{0\})$, which is dense.
Finally, could we weaken second countable to both first countable and weakly Lindelof together? I can adjust the previous example, with a caveat.
Let $X=Y\cup\{\infty\}$ be another altered long ray, where $Y=\omega_1\times[0,1)$ is an open subspace with its lexicographic topology, and each neighborhood of $\infty$ contains $U=\{\infty\}\cup\omega_1 \times (0,1)$.
The proof for weakly Lindelof proceeds similarly as before. This space remains sequentially compact yet not compact, and is now first-countable since $\{U\}$ is a countable base at $\infty$.
The caveat is that this space has very weak separation properties: it's $T_0$, but not even $T_1$.
Hence my question: Is there a first countable, $T_1$, weakly Lindelof, sequentially compact space which is not also compact?
Best Answer
Here is an example:
It is easy to see that $\{[0, \alpha] \setminus F: \alpha < \omega_1, F \subset \omega_1, F \text{ finite}\}$ is a base for a topology on $\omega_1$.
Obviously, this topology is $T_1$. For $\alpha < \omega_1$, $\{[0, \alpha] \setminus F: F \subset [0, \alpha) , F \text{ finite}\}$ is a countable neighborhood base of $\alpha$, hence it is first countable.
It is sequentially compact: Let $(\alpha_n)_{n < \omega}$ be a sequence in $\omega_1$. W.l.o.g, we may assume that $(\alpha_n)_{n < \omega}$ is pairwise distinct, and strictly increasing. Then it converges to its supremum $< \omega_1$.
It is not compact: $\{[0, \alpha]: \alpha < \omega_1\}$ is an open cover without finite subcover.
It is weakly Lindelöf: Let $\mathfrak{U}$ be an open cover of $\omega_1$. For each $n < \omega$ choose $U_n \in \mathfrak{U}$ such that $n \in U_n$. Then $\{U_n: n < \omega\}$ is a countable subcollection of $\mathfrak{U}$, whose union is dense in $\omega_1$.
Remarks
It is easy to see that separable spaces are weakly Lindelöf, and a first countable space is sequentially compact, iff it is countably compact.
Hence, any T1, first countable, separable, countably compact, not compact space provides a counterexample.
There is a rich literature about such spaces, see for instance the article of Peter Nyikos in Open Problems in Topology, 1990, and certainly many more papers published more recently.
In fact, the above example is mentioned in P. Nyikos' article (p. 129). Since $\omega$ is dense in it, it is even separable.
As Nyikos elucidates, the problem becomes much more difficult, if one requires $T_2$ in addition, hence the space will be even regular. There are still a lot of such spaces under additional set-theoretic assumptions. However, as far as I understood this, it was (or still is) not known whether there are examples in ZFC. Might be, it is easier to construct $T_2$ examples, when separability is weakened to weakly Lindelöfness, as in the original question?