Open Filter Characterization of Minimal Hausdorff Spaces For minimal Hausdorff spaces, we have the equivalence -

Hausdorff $X$ is minimal Hausdorff iff every open filter with unique cluster point converges to that point.

I've been able to prove the forward implication, but I've not been able to prove the reverse. Any help would be appreciated!
 A: Suppose $(X, \mathcal{T}')$ is a Hausdorff topology such that $\mathcal{T}'\subseteq \mathcal{T}_X$, where the latter is our original topology on $X$, which has the property that open filters with unique cluster points converge.
Let $x \in X$ and let $\mathcal{F} = \{O \in \mathcal{T}': x \in O\}$ and let $\mathcal{G}$ be the open filter in $\mathcal{T}_X$ generated by $\mathcal{F}$ (which is a filter base in $\mathcal{T}_X$). (Explicitly
$$\mathcal{G} = \{O \in \mathcal{T}_X: \exists O'\in \mathcal{F}: O' \subseteq O\}$$ and it's easy to see this is a filter)
It's clear that $x$ is a cluster point of $\mathcal{G}$ (original topology), and there is no other one: if $x' \neq x$, find disjoint open neighbourhoods $U(x')$ and $U(x)$ in the Hausdorff space $(X, \mathcal{T}')$. Then $U(x') \in \mathcal{T}$ too and it witnesses that $x'$ is not a cluster point of $\mathcal{G}$ as $U(x) \in \mathcal{F} \subseteq \mathcal{G}$.
So $x$ is the unique cluster point of the open filter $\mathcal{G}$ so $\mathcal{G}$ converges to $x$ in $(X,\mathcal{T}_X)$. So every $\mathcal{T}_X$ open set containing $x$ contains a $\mathcal{T}$-open set that contains $x$ (from $\mathcal{F}$ being a filter base for $\mathcal{G}$). As this holds for all $x \in X$, $\mathcal{T}' = \mathcal{T}_X$ (the former has been shown to be a base for the latter, really) and we have a minimal Hausdorff space.
