Countable compactness implies sequential compactness in sequential Hausdorff spaces or Fréchet-Urysohn spaces I am trying to understand the relationship between sequential compactness and countable compactness, but am somewhat confused and not finding answers to my questions online. Would you help me clarify?
Let $X$ be a topological space. We know sequential compactness always implies countable compactness, since otherwise if $\{U_n\}$ was a countable open cover with no finite subcover, define $(x_n)$ such that $x_n \notin \cup_{k=1}^n U_n$. Since $(x_n)$ has a convergent subsequence it follows some $U_n$ must contain infinitely many terms. But this is a contradiction.
In the reverse direction, Wikipedia says if the space is sequential and Hausdorff, the two types of compactness are equivalent. But I couldn't prove this or find a proof.
I was able to prove if $X$ is Fréchet-Urysohn and countably compact then it is sequentially compact. Here is my proof.
Suppose $\exists (x_n)$ with no convergent subsequence. Then $\forall m \in \mathbb{N}, \text{Cl} \{x_n\}_{n \geq m} =\text{SeqCl} \{x_n\}_{n \geq m} = \{x_n\}_{n \geq m}$. So $\{x_n\}_{n \geq m}$ is closed. Now note $U_m = X\backslash \{x_n\}_{n=m}^\infty$ is a countable open cover of $X$ such that $U_1 \subset U_2 \subset \cdots $. By countable compactness $\exists U_N$ which covers $X$. A contradiction.
This made me wonder if sequential Hausdorfness implies a space is Fréchet-Urysohn, and Wikipedia's statement would follow, but I couldn't find a proof of this.

Your proof idea works with minor modification even if $X$ is just sequential and $T_1$, rather than Frechet-Urysohn.  Indeed, if $(x_n)$ has no convergent subsequence, this means that no sequence in the set $\{x_n\}$ that takes infinitely many values can have a limit, since such a sequence would then have a subsequence which is also a subsequence of $(x_n)$.  If $X$ is $T_1$ (so a convergent sequence that takes only finitely many values can only have one of those values as its limit), this implies that for any $m$, the set $\{x_n\}_{n\geq m}$ is sequentially closed and thus closed since $X$ is sequential, and then the rest of your proof goes through.
(With a bit more work, the $T_1$ assumption can in fact be dropped: instead of considering the sets $\{x_n\}_{n\geq m}$, consider $\bigcup_{n\geq m}\overline{\{x_n\}}$, and again the assumption that $(x_n)$ has no convergent subsequence can be used to show these sets are all sequentially closed, and also that their intersection as $m\to\infty$ is empty.)
