Past exam of the topology course of the University of Oxford / 2023 / 1 / b / iii

The answer is no, because the number of non-isolated points differs. In $X$, every $1/n$ is an isolated point: a sufficiently small open interval around it contains no other member of $X$. The only non-isolated point is $0$, since every neighbourhood of $0$ contains some $1/n$.

In the product topology, $(x,y)\in X\times X$ is an isolated point exactly when both coordinates are isolated points of $X$. One direction follows because the product of two open singletons is open. For the other, any basic open rectangle containing $(x,y)$ contains another point if either coordinate has no open singleton. Hence the set of non-isolated points of $X\times X$ iswhich is infinite. A homeomorphism takes open singletons to open singletons, so it induces a bijection between the sets of non-isolated points. A space with just one such point cannot be homeomorphic to a space with infinitely many.