Converting between $T_1$ and $T_2$ Given that every $T_2$ space is a $T_1$ space, is it possible to start with a $T_1$ space and to specify in terms of its sets a family of additional sets sufficient to make that $T_1$ space into a $T_2$ space? If so, can this be done for any $T_1$ space or only particular ones?
 A: Here’s an example of a $T_1$ space whose topology $\tau$ has no minimal Hausdorff extension that bears much resemblance to $\tau$ and that has at least two minimal Hausdorff extensions that are not comparable in the lattice of topologies on the underlying set.
Let $\mathscr{E}$ be the usual topology on $\Bbb Q$. Let $\mathscr{E}_0=\{U\in\mathscr{E}:0,1\notin U\}$ and
$$\mathscr{B}=\mathscr{E}_0\cup\{\Bbb Q\setminus F:F\subseteq\Bbb Q\setminus\{0\}\text{ is finite}\}\cup\{\Bbb Q\setminus F:F\subseteq\Bbb Q\setminus\{1\}\text{ is finite}\}\;;$$
$\mathscr{B}$ is a base for a $T_1$ topology $\tau$ on $\Bbb Q$. $\langle\Bbb Q,\tau\rangle$ is not Hausdorff, however, since $0\in\operatorname{cl}U$ for every non-empty $U\in\tau$. Clearly $\tau\subseteq\mathscr{E}$. Suppose that $\tau'$ is a Hausdorff topology on $\Bbb Q$ such that $\tau\subseteq\tau'\subseteq\mathscr{E}$. In ‘$T_\nu$-Abgeschlossenheit und $T_\nu$-Minimalität’, Mathematische Zeitschrift $88$ $(1965)$, $285$-$294$, Horst Herrlich showed that $\mathscr{E}$ contains no coarser minimal Hausdorff topology, and it follows immediately that the same is true of $\tau'$. In particular, $\tau'$ is not minimal Hausdorff.
It’s still possible, of course, that $\tau$ can be extended to a minimal Hausdorff topology that is not contained in $\mathscr{E}$. However, Theorem $5.2$ of Jack Porter & Johannes Vermeer, ‘Spaces with coarser minimal Hausdorff topologies’, Trans. Amer. Math. Soc., Vol. $289$, Nr. $1$, $59$-$71$, says that a countable Hausdorff space that has a coarser minimal Hausdorff topology is scattered. Since $\Bbb Q$ is countable, this means that if $\tau'\supseteq\tau$ is a minimal Hausdorff topology, then $\langle\Bbb Q,\tau'\rangle$ is a scattered space and therefore bears very little resemblance to $\langle\Bbb Q,\tau\rangle$.
For instance, we can construct $\tau'$ from $\tau$ by isolating every point of $\Bbb Q\setminus\{0\}$; then $\langle\Bbb Q,\tau'\rangle$ is compact Hausdorff and hence minimal Hausdorff. But we can just as well construct $\tau''$ from $\tau$ by isolating every point of $\Bbb Q\setminus\{1\}$: it’s minimal Hausdorff for the same reason, and it’s not comparable with $\tau'$ in the lattice of topologies on $\Bbb Q$.
A: I suspect that every way of doing this is somehow either arbitrary or trivial.
Suppose we have an operation $\mathcal{H}$, such that if $(X, T)$ is a T1 space
then $T \subset \mathcal{H}(T)$ and $(X, \mathcal{H}(T))$ is a T2 space.
Suppose furthermore that this operation preserves symmetry, in the sense that if
$f: X \to X$ is a homeomorphism of $(X, T)$ to itself, then it is also a homeomorphism of
$(X, \mathcal{H}(T))$ to itself.
Finally, suppose that the operation is monotone, in the sense that if $T \subset T'$, then
$\mathcal{H}(T) \subset \mathcal{H}(T')$.
With these, apparently reasonable, assumptions it is already unavoidable that $(X, \mathcal{H}(T))$
is discrete, whatever $T$ is. To see this, consider the cofinite topology. It is T1
and every bijection is a homeomorphism. The only T2 topology with the property
that every bijection is a homeomorphism is the discrete one.  Since the cofinite topology
is the coarsest T1 topology and the discrete topology is the finest of all,
monotonicity then demands that $\mathcal{H}$ makes every T1 topology discrete.
