How can you have a non-regular Hausdorff space? Both definitions seem to be identical, 2 points with disjoint neighborhoods.
Pls don't point out that this is not a formal definition. It is a simple (but incomplete) description that makes assumptions.
Note that the other time this was asked, the replies immediately decayed into an argument about terminology. Another answer failed to distinguish regular space from completely regular.
Finally, I'm sorry for all the disclaimers, but I'm getting mod warnings and question deletions, and I'm not sure how to comply.
In Hausdorff spaces $X$ we have disjoint neighbourhoods for two points $x \neq y$ in $X$ and in regular spaces we have disjoint neighbourhoods for any pair $x,C$ where $x \notin C$ and $C$ is closed.
So it's much harder to satisfy regularity because we need open neighbourhoods for all closed sets and points not in the closed set, and there will in general (certainly for $T_1$ spaces, where sets like $\{y\}$ are closed too, so there regularity implies Hausdorffness) by more of such point-closed set pairs than distinct point pairs.
So more conditions to fulfil, so we expect non-regular Hausdorff spaces to exist a priori, and they do. Not many you'll meet in practice, as there most spaces are even completely regular even (in analysis) or even metric or ordered (which satisfy even better conditions, like normality).