An infinite-dimensional pre-Hilbert space not homeomorphic to its own square

Given an arbitrary infinite-dimensional separable complete linear metric space $X$, there exists a direct sum decomposition $X = {V_0} \oplus {V_1}$ such that each summand ${V_i}$ intersects every linearly independent Cantor set in $X$ (this decomposition can be considered as a linear analogue to the classical Bernstein's decomposition into totally imperfect sets).

Theorem. Each summand $V$ of such a decomposition is not homeomorphic to its own square, and if $T:V \to V$ is a linear bounded operator, then either the kernel or the range of $T$ is finite-dimensional.

In the case of $X = {l_2}$ this provides an example of a space $V$ with the properties stated in the title, which answers a well-known question, cf. Arhangelskiĭ [A, Problem 21] and Geoghegan [G, Problem (LS 12)].