Let $X$ be a topological (Hausdorff) space and let $(X_\alpha)_\alpha$ be a directed family of subsets. We say that $(X_\alpha)_\alpha$ generates the topology of $X$ if a subset $U \subseteq X$ is open iff $U\cap X_\alpha$ is open in $X_\alpha$ with respect to the induced topology. Another way of saying this is that $X$ is the direct limit of the topological spaces $(X_\alpha)_\alpha$ where each $X_\alpha$ holds the subspace topology.

It is well-known that the topology of a metrizable space is generated by the family of all compact subsets (one says $X$ is a "$k$-space") since the topology is determined by convergent sequences.

With exactly the same argument, we obtain the result that the topology of a metrizable space is generated by all countable subsets.

My question is now:

For wich non-metrizable Hausdorff spaces it is true that the topology is generated by countable subsets?

Trivially, this also holds for countable spaces but I would expect that there are many more examples. Also, a counter-example would be interesting.

In particular I am interested in examples of the form

$\mathbb R^I := \prod_{i\in I} \mathbb R$ or $(\mathbb Z/2\mathbb Z)^I := \prod_{i\in I} (\mathbb Z/2\mathbb Z) $

for an uncountable index set $I$, but of course all other examples or counter-examples are welcome as well.

Thanks in advance, Tom