Let $(X,\tau)$ be a Hausdorff space. Let $[X]^2 = \big\{\{x,y\}: x,y\in X \land x\neq y\big\}$. For $U,V\in \tau$ with $U\cap V = \emptyset$ we set $[U,V] = \big\{\{x,y\} \in [X]^2: x\in U\land y\in V\big\}$.
We endow $[X]^2$ with the topology $[\tau]^2$, which is generated by $\{[U,V]: U,V\in \tau\land U\cap V =\emptyset\}$.
Is there a non-discrete Hausdorff space $X$ with more than 3 points such that $X\cong [X]^2$?
It is easy to see that the operation $X \mapsto [X]^2$ preserves the properties: countable, second-countable, regular, no isolated points. Hence $\mathbb{Q} \cong [\mathbb{Q}]^2 $.
Write $F(X)=[X]^2$. Then $F$ is an endofunctor on the category of Hausdorff spaces and inclusions of clopen subspaces, and it preserves filtered colimits. Given any Hausdorff space $X$ with an inclusion of $X$ as a clopen subspace of $F(X)$, it follows (as in this answer) that $F^\omega(X)=\operatorname{colim} F^n(X)$ satisfies $F(F^\omega(X))\cong F^\omega(X)$. It follows easily that every Hausdorff space embeds in a fixed point of $F$ (for instance, given any $X$, if $Y=X\coprod\mathbb{N}$ then $Y$ is naturally homeomorphic to a clopen subspace of $F(Y)$ and then $X$ embeds in $F^\omega(Y)$).
(The restriction to inclusions of clopen subspaces is probably stronger than is needed here; the key point is that you restrict to a nice enough class of inclusions such that finite products commute with sequential colimits of such inclusions.)
Yet another example: $X:=\{−3,−2,−1\}∪\mathbb{Q}_+$ with the standard Euclidean topology. $\mathbb{Q}_+$ denotes the positive rationals.
I post it on an explicit request in the comments, but this is in fact a combination of the answer by Ramiro de la Vega, the finite example banned by Dominic van der Zypen and the fact that a product of a finite set by the rationals is homeomorphic to the rationals.
