A space $X$ is called Corson compact if there is a set $I$ such that $X$ is homeomorphic to a compact subspace of $\{x \in \mathbb{R}^I:\{i\in I: x(i) \neq 0\} \text{ is countable }\}$, $\mathbb{R}^I$ being eqquipped with the product topology.
I am looking for an example of Corson compact. Is true that $\mathbb{N}$ with the co-finite topology is Corson compact?
All Corson compact spaces are completely regular (as a subspace of the space $\mathbb{R}^I$), so the cofinite topology is not.
As to examples: it's classical that all compact metric spaces are Corson compact by the Tychonoff embedding theorem, so $[0,1]$ will do, e.g.
for much more info see this blog post, e.g.