Every sequentially compact space is countably compact

Every sequentially compact space is countably compact.

The most that I can get out of this $(\Rightarrow)$ is that ${x_{n_j}} \subseteq \bigcup_{i=1}^{n_0}O_{x_i}$, where $O_{x_i}$ is open neighborhoods of the finite sequence points and $O_{x_{n_0}}$ is the open set containing the infinitely many points beyond $x_{n_0}.$ I'm a bit stuck beyond this. I tried relating this to: $X$ is countably compact iff the intersection of every sequence of nonempty nested closed proper subsets of $X$ is nonempty, but I couldn't figure out how to show a contradiction.

Anyone have any ideas?

• You’ll find a proof in the statement of this question; ignore the title to the question. – Brian M. Scott Jun 7 '16 at 22:13

• Is $A$ the topological space? – Oliver G Jun 8 '16 at 2:04
Use the fact that $X$ is countably compact iff every infinite set $A$ has a point $p$ such that for every open neighbourhood $O$ of $p$, $O \cap A$ is infinite.
For a proof, see your own question here. Such a $p$ is called an ($\omega$-)accumulation point of $A$.
So if $A$ is infinite, pick $a_n, n \in \mathbb{N}$, all different, in $A$. This defines a sequence $(a_n)_n$ in $X$, so it has a convergent subsequence $(a_{n_k})_k$ by sequential compactness of $X$, and say that $p$ is a limit of this convergent subsequence.
Now if $O$ is any open neighbourhood of $p$, it will contain all $a_{n_k}$ for $k \ge K_0$ for some $K_0 \in \mathbb{N}$, by convergence of the subsequence. In particular, $O \cap A$ is infinite. The above characterisation shows we are done: $X$ is countably compact, as we can find an accumulation point $p$ for every infinite $A \subseteq X$.