Is Every Sequentially Compact, First Countable Space Also Compact? I want to know if sequential compactness in first countable space implies compactness. I know the implication is correct in metric space (a proof that uses Lebesgue number), but I cannot mimic a proof in a general first countable space.
Or perhaps we cannot get compactness when we only have sequential compactness, but only countable compactness or Lindelof?

Firstly, sequential compactness is stronger than countable compactness, and spaces like $[0,1]^\mathbb{R}$ (in the product topology) are compact (so countably compact as well) but not sequentially compact. And certainly sequences behave better in first countable spaces, because closures and continuity can be characterised using sequences, just as in the metric case.
But the hypothesis is false: the classical example is to take the $[0,\omega_1)$, all ordinals below the first uncountable ordinal $\omega_1$, in the order topology. This is first countable (it's locally metrisable, even), sequentially compact and countably compact, but not compact. This settles the sequentially compact (and countably compact) case.
Lindelöf doesn't help at all to get compactness, even in metric spaces (reals, etc.), so I don't understand the last remark.
What is true, is that for first countable spaces, $X$ sequentially compact iff $X$ countably compact (see e.g. here) but we cannot go beyond countably compact. The Lindelöf part of compactness has to be assumed as well, one could say.
It can be interesting to find weaker conditions than Lindelöf, so find conditions "P" such that for $X$ that are first countable and sequentially compact: $X$ has "P" iff $X$ is compact. Of course, taking "P" as Lindelöf works, but maybe weaker ones will also work? Maybe weaker covering properties?
