Is it true that $\displaystyle \mathbb{|R|} < \mathbb{|R}^2|$
No. $\displaystyle \mathbb{R}$ is equinumerous with (0,1); the bijection between these sets can be constructed using the tangent function on $\displaystyle (-\pi/2,\pi/2)$. Similarly, $\displaystyle \mathbb{R}^2$ is equinumerous with $\displaystyle (0,1)\times(0,1)$. Now, there is a bijection $\displaystyle (0,1)\to(0,1)\times(0,1)$. See section 4 in this PDF document.