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?
 A: 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 $A \subseteq X$.   
A: The finite case is trivial. Consider ${N(x_1),N(x_2),N(x_3),...}$ an arbitrary countable open cover. ${x_1,x_2,x_3,...}$ partitions into finitely many subse uences all of which are convergent. Without loss of generality ${x_1,x_2,x_3,...} $converges to $x_\infty$. But $x_\infty$ is in $N(x_k)$ for some $k$. It follows that we have a finite subcover.
A: Let $X=\bigcup\limits_n V_n$ and suppose there is no finite subcover. Then for all $n$ we have $V_1\cup V_2\cup\dots\cup V_n\subsetneq X$. So pick $x_n\in X\setminus V_1\cup V_2\cup\dots\cup\ V_n$. Then $(x_n)$ is a sequence satisfying $x_n\notin V_m$ for all $m\leq n$. Let $(x_{n_k})$ be a convergent subsequence converging to some $x$. There is some $V_N$ containing $x$. So all but finitely many terms of $(x_{n_k})$ lie in $V_N$ which is a contradiction.
