Showing $(0,1)^3$ is not homeomorphic to $[0,1)^3$
So, if these spaces are homeomorphic, then if we remove any single point from each of them, they will still be homeomorphic.
$(0,1)^3 - \{p\}\cong S^2$
While
$[0,1)^3 - \{0\} \cong \mathbb{R}^3$
But $H_2(S^2) \cong \mathbb{Z}$ and $H_2(\mathbb{R}^3) \cong 0$.
Thus we have $(0,1)^3 - \{p\} \ncong [0,1)^3 - \{0\}$
$\therefore$ $(0,1)^3$ $\ncong$ $[0,1)^3$