A sequentially compact space is countably compact.

Figure 3 on page 20 of Counterexamples in Topology, Steen and Seebach has some assumption? (like $$T_1$$, first countable) One of the 6 implications shown on the picture is

A sequentially compact space is countably compact.

I'm stuck with the proof and before asking for a direct help, I would like to be sure about the correctness of my interpretation: no assumption. • As for the proof, try the contrapositive. Assume $X$ isn't countably compact and construct a sequence without convergent subsequence. Nov 18 '19 at 20:59

2. Every infinite subset of $$X$$ has an $$\omega$$-accumulation point in $$X$$. (i.e. $$p$$ such that every neighbourhood contains infinitely many points of that set.)
3. Every sequence $$(x_n)_n$$ in $$X$$ has an accumulation point in $$X$$. (i.e. a $$p$$ such that for every neighbourhood $$O$$ of $$p$$ and every $$n$$ there is some $$m \ge n$$ such that $$x_m \in O$$.)
4. Every countable family of closed subsets of $$X$$ with empty intersection has a finite subfamily with empty intersection.
And it's immediate that $$\text{CC}_3$$ is implied by sequential compactness, as the limit $$p$$ of a subsequence (which then exists) is always an accumulation point of the sequence, regardless of extra conditions, just definitions. This justifies their arrow.
Right: there are no extra assumptions. Take the double pointed countable complement topology (example 21 of their book). It's not even $$T_0$$, but, for them, it is weakly countably compact.