Topological spaces with isomorphic categories of open subsets Given two topological spaces $X$ and $Y$, let $\text{Op}(X)$ (resp. $\text{Op}(Y)$) be the category of open subsets of $X$ (resp. $Y$). If the two categories $\text{Op}(X)$ and $\text{Op}(Y)$ are isomorphic, under what conditions can we say that $X$ and $Y$ are also isomorphic? What kind of information of $X$ can we learn from $\text{Op}(X)$? More importantly, what if $X$ is a quasi-projective variety (Zarski topology)?
 A: This doesn't answer your question entirely but I hope it will help.
The idea to consider categories of opens instead of points of spaces leads to the so-called locales. A locale is a poset with finite intersections and all unions (products and coproducts), which is distributive. Morphisms of locales are morphisms that commute with all that structure, except that we take an opposite category, because a map of posets $\mathcal O (X) \to \mathcal O (Y)$ corresponds to the map $Y \to X$ in the opposite direction.
Now, a point of a locale $L$ is a map $\{0,1\} \to L$ (because it's a map from the final object, like in $\text{Top}$). A point of a space $p \in X$ determines a point $\bar{p}$ of a locale $\mathcal O (X)$ as a map of posets $\mathcal O (X) \to \{0,1\}$ such that it equals $0$ on those subsets $V \subset X$ that do not contain $x$.  When the opposite is true, the space is called sober. More precisely, a sober space is a space $X$ such that for all $P \in O (X)$ satisfying
$P \not = X$
and $U \cap V \subset P \implies U \subset P \text{ or } V \subset P$
[here we are inspired by the properties of those points $\bar{p}$ of $\mathcal O (X)$ that come from $p \in X$], there is a unique point $x$ such that $P$ in the complement of the closure of $x$.
Points of locales can be topologized, which gives rise to the adjunction
$\text{loc}: \text{Top}\substack{\rightarrow \\[-0.1ex] \leftarrow} \text{Loc}: \text{pt}.$
A space is homeomorphic to points of its locale (via the unit of this adjunction) precisely when it's sober.
