PreScriptum. Having almost finished the answer below I saw the answer by Simon Henry which largely subsumes mine. I still decided to post it as it contains some details/proofs, so may be viewed as an addendum to that answer.
Concerning 2.:
First note that $f:X\to Y$ is an open embedding in your sense iff $X$ is up to isomorphism an open of $Y$.
(More precisely this means that there is an $U_X\in\operatorname{Opens}(Y)$ and an isomorphism $i:\operatorname{Opens}(X)\cong\{U\in\operatorname{Opens}(Y)\mid U\subseteq U_X\}$ such that $if^*(\_)=U_X\cap\_$ for the frame homomorphism $f^*:\operatorname{Opens}(Y)\to\operatorname{Opens}(X)$ determining $f$. Proof - if the latter holds then the needed map $Y\to$ Sierpiński corresponds to the frame homomorphism $\{\varnothing,\{o\},\{o,c\}\}\to\operatorname{Opens}(Y)$ sending $\{o\}$ to $U_X$, since the pushout of this map along the embedding of the open point into Sierpiński is the quotient of $\operatorname{Opens}(Y)$ by the smallest frame congruence identifying $U_X$ and $Y$; conversely, given such a map $f:Y\to$ Sierpiński, take $U_X=f^{-1}(o)$.)
Next observe that if $\operatorname{Opens}(Y)=\operatorname{Powerset}(S)$, then for any subset $T\subseteq S$ the frame homomorphism $T\cap\_:\operatorname{Powerset}(S)\to\operatorname{Powerset}(T)$ determines an open embedding in $Y$ of an $X$ with $\operatorname{Opens}(X)=\operatorname{Powerset}(T)$.
This in particular applies to $T\subseteq T\times T$ and we are done given that for $\operatorname{Opens}(X)=\operatorname{Powerset}(T)$ and $\operatorname{Opens}(X')=\operatorname{Powerset}(T')$ one has $\operatorname{Opens}(X\times X')=\operatorname{Powerset}(T\times T')$.
For the converse direction (and this also addresses 1. a bit), there is an additional necessary condition: if $\operatorname{Opens}(X)=\operatorname{Powerset}(T)$, then not only the diagonal $X\to X\times X$ but also $X\to\text{point}$ is open (i. e. the image of any open of $X$ under $X\to\text{point}$ is an open subset of the single point locale). Such locales are usually called $\textit{overt}$.
If the logic is not classical, there might exist non-overt locales with open diagonal. For example, if a singleton $\{*\}$ has a non-complemented subset $S\subset\{*\}$, then $S$ determines an open sublocale $U$ of the single point locale with $\operatorname{Opens}(U)=\operatorname{Powerset}(S)$; then this open sublocale has the complementary closed sublocale $C$, and if $C$ would be open too then $S$ would be complemented in $\{*\}$. Now $C$ is an example of a non-overt locale with open diagonal since ($C$ being a sublocale of the single point locale) the image of $C\to\text{point}$ is $C$ and the diagonal $C\to C\times C$ is an isomorphism.
In particular, it follows that there is no set $T$ with $\operatorname{Opens}(C)$ isomorphic to $\operatorname{Powerset}(T)$, since otherwise $C$ would be overt. (Actually $\operatorname{Opens}(C)$ is isomorphic to $\{S'\subseteq\{*\}\mid S'\supseteq S\}$.)