Under "Topology of sequentially open sets" section of the Wikipedia page Sequential Space, there is a claim which says
any sequentially Hausdorff(i.e. every convergent sequence has a unique limit) space $(X,\tau)$, its sequential topology $\bar\tau=\{S\subset X|\text{SeqInt}_\tau S=S\}$ is Hausdorff(the space $(X,\bar\tau)$ is Hausdorff).
Proving the sequentially Hausdorffness of a space that has Hausdorff sequential topology is rather easy, since any sequence has the same convergence under its original topology and its sequential topology.
The trouble occurs to me when I try to prove the original claim. I transformed the original claim into proving $\exists p,q\in X(\forall U(\ni p)\in\bar\tau,q\in \text{Cl}_{\bar\tau}U)$ being false, my idea is to try constructing a sequence that converges to both $p,q$ under this condition, but I am unable to progress further.
Several examples(of co-countable topology) suggest me these $T_2$ non-separable(but $T_1$ separable) points would obtain discrete topology, does this still hold on other examples?
Best Answer
The Wikipedia page you cited actually defines $(X,\tau)$ to be sequentially Hausdorff iff $(X, \bar{\tau})$ is Hausdorff (in the usual sense).
The property that convergent sequences have unique limits (sometimes called US) is quite different, so the remark "sequentially Hausdorff(i.e. every convergent sequence has a unique limit)" is false. I cannot find it in that form on the page either (I haven't looked in the edits, so it might have been there in the past and been corrected).
In fact example 5.3 in Franklin's classic paper has an example of a countable, compact, sequential space $M_1$ that has unique sequential limits but which is not Hausdorff. So $M_1$ is not sequentially Hausdorff, trivially (as for it, $\tau=\bar{\tau}$, being sequential).
So fix your idea of sequentially Hausdorff means. Your initial claim is false as it stands.. But it's true by definition with the correct definition.