Can a space $X$ be homeomorphic to its twofold product with itself, $X \times X$? Let $X$ be  a  topological  space  of  infinite  cardinality. Is  it  possible  for  any  $X$  to  be  homeomorphic  to  $X\times X$  $?$
For example, $\mathbb R$  is  not  homeomorphic  to  $\mathbb R^{2}$, and $S^{1}$  is not  homeomorphic  to  $S^{1} \times S^{1}$ . What  other  topological  spaces  might we consider$?$ What  properties  of  a  space  may  ensure  or  contradict  this  possibility$?$  From  the  little  topology  I  have  learnt  yet,  I  have  not  seen  this  happening.

Yes, consider $X := \Bbb Z$ endowed with the discrete topology.
For any topological manifold $M$, $\dim (M \times M) = \dim M + \dim M = 2 \dim M$. Since the dimension of a nonempty topological manifold is well-defined, there is no positive-dimensional topological manifold $M$ for which $M \cong M \times M$, which in particular excludes $R$ and $S^1$ as observed. This implies that the example $X = \Bbb Z$ is the only example that is a (second countable) topological manifold.
