Suppose $\mathcal{X}=(X,\tau)$ is a non-Hausdorff space. There are two ways to make precise "the sets that need to become open if we refine $\mathcal{X}$ to become Hausdorff" that I'm interested in comparing:
Most obviously, we can just take the intersection of all Hausdorff refinements. Call this $\eta^-(\tau)$.
Alternatively, we can restrict attention to Hausdorff refinements which preserve continuity of functions; that is, we take the intersection of all Hausdorff topologies $\rho$ on $X$ with $\rho\supseteq \tau$ and with the property that every $\tau$-continuous function $f:X\rightarrow X$ is also $\rho$-continuous. Call this $\eta^+(\tau)$.
(The former is much simpler to think about, but the latter feels better - to me at least - insofar as we think about the "structure" of a topological space consisting not just of its open sets but also its monoid of continuous endofunctions, which we sometimes do.)
We always have that $\eta^+(\tau)$ refines $\eta^-(\tau)$ (intersections over "larger" families are "smaller"). I'm interested in how far apart they can be. In particular, the following seems like a good test question:
Is there a $T_1$ space $\mathcal{X}$ such that $\eta^+(\tau)$ is Hausdorff but $\eta^-(\tau)$ is not Hausdorff?
(We need $T_1$-ness to avoid a silly answer: if $\mathcal{X}$ is indiscrete, then the only "continuous-function-preserving" proper refinement of $\tau$ is the discrete topology. In particular, if $\mathcal{X}$ is infinite and indiscrete then $\eta^+(\tau)$ is the discrete topology and $\eta^-(\tau)$ is the cofinite topology. Thanks to Eric Wofsey for pointing this out!)
If $X$ is infinite and $\tau$ is cofinite, then $\eta^-(\tau)=\tau$ (for instance, consider the intersection of all topologies that make $X$ the 1-point compactification of a discrete space) is not Hausdorff. However, $\eta^+(\tau)$ is discrete, since every bijection $X\to X$ is $\tau$-continuous and the only Hausdorff topology with this property is discrete (any Hausdorff topology on $X$ must have a nonempty open set whose complement has cardinality $|X|$ and then by permuting and intersecting two such sets you can get any singleton).