$T_2$-spaces where all non-empty open sets are homeomorphic We say that a $T_2$-space $(X,\tau)$ has homeomorphic open sets if every non-empty open set $U\subseteq X$ endowed with the subspace topology is homeomorphic to $(X,\tau)$.
The rationals with the Euclidean topology are an example of a space with homemorphic open sets, as well as $\{0,1\}^\lambda$, where $\lambda$ is an infinite cardinal and $\{0,1\}$ carries the discrete topology.
Is there for every infinite cardinal $\kappa$ a space of cardinality $\kappa$ with homeomorphic open sets?

Edit. Apologies for falsely claiming that $\{0,1\}^\lambda$ is a space with homeomorphic open sets - Joel David Hamkins' comment below gives an argument refuting my claim.
 A: The Baire space $\omega^\omega$ has homeomorphic open sets: it is an immediate consequence of  the fact that it is the unique nonempty zero-dimensional Polish space, up to homeomorphism, every compact subset of whose has empty interior (theorem 7.7 in Kechris). In fact we actually have better:

Proposition. For every infinite set $X$ with the discrete topology, $X^\omega$ has homeomorphic open sets.

Proof. Let $U \subseteq X$ be nonempty open. I claim that $U$ is a union of exactly $\kappa$ disjoint cones, where $\kappa = |X|$. To see this, it is enough to show that $U$ is a union of $\leqslant \kappa$ disjoint cones, as a cone is itself a union of $\kappa$ disjoint cones. But if $T \subseteq X^{< \omega}$ is a nonempty tree such that $[T] = U^c$, then $U$ is the disjoint union of all cones $N_s$ where $s \notin T$ but $s_{\restriction |s| - 1} \in T$.
Now, writing $U$ as a union of $\kappa$ disjoint cones, we can get a homeomorphism between $U$ and $X$ by gluing together homeomorphism between these cones and cones of the form $N_{(x)}$, $x \in X$.
So the answer to Dominic's question is positive at least for cardinals of the form $\kappa^\omega$.
Edit: Will Brian remarked, in the comments, that the same argument actually enables to show that the subspace of $X^\omega$ whose elements are eventually constant also had homeomorphic open sets. So for every infinite cardinal, there exists a space with homeomorphic open sets.
