= Solution
The answer is \b[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 $X\times X$ is
$$
(\{0\}\times X)\cup(X\times\{0\}),
$$
which 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.
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