Counterexample to make the difference between weakly sequentially compact and sequentially compact explicit

Question

Some background (all of this is from Barry Simon's "Real Analysis - A comprehensive course in analysis - vol 1"):

Reading both definitions I can see the difference between weakly sequentially compact and sequentially compact, but I can't find an example of a topological space which is weakly sequentially compact but not sequentially compact or at least one space where we have a sequence with at least one limit point but which doesn't have any convergent subsequence.

Here are some potential counter examples some friends and I thought of but didn't work:

Consider $\mathbb{N} \times \{0, 1\}$ where $\mathbb{N}$ has the discrete topology and the latter the indiscrete one. Now, the sequence:

$(1, 0), (1, 1), (2, 0), (2, 1), \cdots$

clearly has no limit point. It's also pretty clear that it doesn't have any convergent subsequence.

Answer 1

$\{0,1\}^{\mathbb{R}}$ is weakly sequentially compact (it's even compact Hausdorff which implies that property), but it has sequences without convergent subsequences.

$\beta \omega$, the Cech-Stone compactification of the countable discrete space $\omega$ is another example, for the same reasons. The sequence $x_n = n \in \omega$ is a sequence without a convergent subsequence.

"Weakly sequentially compact" is a confusing name IMHO, just call it countably compact (every countable open cover has a finite subcover), to which it is equivalent. I show this fact in my answer here.

Answer 2

I would try the interval $[-1,1]$ with two origins. The sequence $\{1/n\}$ has two limit points (the two origins), but no convergent subsequence.