My favourite counterexample is purely academic: it does not have any applications, but I think it is pretty.
Let $X = \mathbb{N} \times \mathbb{N}$. Define a non-empty set $U \subseteq X$ to be open if for cofinitely many $x \in \mathbb{N}$ the set ${ \{ y \in \mathbb{N} \vert (x,y) \in U}$ U\}$ is cofinite.
Construct a sequence in $X$ that hits every point in $X$ exactly once. In other words, take a bijection $\mathbb{N} \rightarrow X$. Then:
- $X$ is countable;
- every point in $X$ is an accumulation point of this sequence, but
- the sequence has no convergent subsequences.
In particular, this is an example in a countable set that accumulation point of a sequence does not have to be a limit of a subsequence. I call this the Herreshoff topology for the (high-school) student of mine who came up with it. (I could not find it anywhere else, although I do not discard that I did not look hard enough.)
