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 $\kappa^\omega$.

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.

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 nonempty 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 $\kappa^\omega$.

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 $\kappa^\omega$.