Prove that the image of the bilinear application $$ \mathbb{R}^2\times\mathbb{R}^2 \ni \big((x_1,x_2),(y_1,y_2)\big) \mapsto \varphi \big((x_1,x_2),(y_1,y_2)\big) = (x_1y_1,x_1y_2,x_2y_1,x_2y_2)\in\mathbb{R}^4 $$ is not a vector space.
My work. If $\mathrm{Im}(\varphi)\subseteq\mathbb{R}^4$ is a subspace of $ \mathbb{R}^{4}$ then for any scalars $\mu,\lambda\in\mathbb{R}$ and vectors $$ p=(a\cdot c, a\cdot d, b\cdot c, b\cdot d) \qquad \mbox{ and } \qquad q=(u\cdot z, u \cdot w, v\cdot z, v\cdot w) $$ in $\mathrm{Im}(\varphi)$ we have $$ \mu\cdot p+\lambda\cdot q \in \mathrm{Im}(\varphi). $$ Therefore, if we can show that there are scalars $\lambda,\mu\in\mathbb{R}$ and vectors $p,q\in \mathrm{Im}(\varphi)$ such that $$ \mu\cdot p+\lambda\cdot q \notin \mathrm{Im}(\varphi) $$ then $\mathrm{Im}(\varphi)$ will not be a vector space. The relationship of non-pertinence above is equivalent to the following statement. For all vector $(x_1y_2,x_1y_2,x_2y_2,x_2y_2)\in\mathrm{Im}\varphi$ the system below has no solution $$ \begin{array}{rcl} \mu a c+ \lambda u z &=& x_1y_1\\ \mu a d+ \lambda u w &=& x_1y_2\\ \mu b c+ \lambda v z &=& x_2y_1\\ \mu b d+ \lambda v w &=& x_2y_2\\ \end{array} $$ But I have been unable to find real numbers $a$, $b$, $c$ and $d$ and scalars $\lambda$ and $\mu$ for which the above system has no solution for $x_1$, $x_2$, $y_1$ and $y_2$ varying in $\mathbb{R} $.
Your claim is true as $\,\operatorname{Im}\varphi\,$ is $\,$ not closed $\,$ under addition:
$$\varphi\big((1,0),(1,0)\big) \:=\: (1,0,0,0)\\ \varphi\big((0,1),(0,1)\big) \:=\: (0,0,0,1)$$
Now consider the sum $(1,0,0,1)$ of both ...
Can it be contained in $\operatorname{Im}\varphi$?