Image of a subset of $\Bbb{R}^2$ is open 
Let $f:\Bbb{R}^2 \to \Bbb{R}^2$ be continuously differentiable. Suppose $f'(0)$ has non-zero determinant. Let $U = \left\{x \in \Bbb{R}^2 : ||f'(x)-f'(0)||<\frac{1}{2||f'(0)||}\right\}$. Show that $f(U)$ is open.

I have tried doing this using generic properties of a norm but that didn't work. I suspect that I'll need to specifically use what the norm is. And I don't know what that is. The wikipedia article on norms for matrices didn't help.
So an answer telling what the norm is( in this case for a $2\times2$ matrix) and the general direction to proceed in would be helpful. 
 A: This is not true. Take $f(x,y)=\left(\left(x-\frac14\right)^2,\left(y-\frac14\right)^2\right)$. Then $f'\left(0,0\right)=-\frac12\operatorname{Id}$ and so $\left\|f'\left(0,0\right)\right\|=\frac12$. On the other hand, $f'\left(\frac14,\frac14\right)=0$ and so $\left(\frac14,\frac14\right)\in U$. But then $f(U)$ is not open, since $(0,0)=f\left(\frac14,\frac14\right)\in U$ but both coordinates of any element of $f(U)$ is greater than or equal to $0$.
A: Your statement is in general not true. Take
$$ f_c: \mathbb{R}^2 \rightarrow \mathbb{R}^2, \ f_c(x,y)=\left(\sin \left(\frac{x}{c}\right), \sin\left(\frac{y}{c}\right)\right).$$
One computes
$$ f_c'(x,y)= \frac{1}{c}\begin{pmatrix} \cos\left(\frac{x}{c}\right) & 0 \\ 0 & \cos\left(\frac{y}{c} \right) \end{pmatrix} = \frac{1}{c} f_1'\left( \frac{x}{c}, \frac{y}{c} \right).$$
We have
$$ \Vert f_c'(0,0) - f_c'(x,y) \Vert 
\leq 2 \max_{(u,v)\in [0, 2\pi c]\times [0,2\pi c]} \Vert f_c'(u,v) \Vert 
= \frac{2}{c} \max_{(u,v)\in [0, 2\pi]\times [0,2\pi]} \Vert f_1'(u,v) \Vert.$$
Thus, if we choose $c>0$ such that $ c^2> 2 \Vert f_1'(0,0) \Vert \cdot \max_{(u,v)\in [0, 2\pi]\times [0,2\pi]} \Vert f_1'(u,v) \Vert $, then
$$ U= \{ (x,y)\in \mathbb{R}^2 : \ \Vert f_c'(x,y) - f_c'(0,0) \Vert < \frac{1}{2\Vert f'(0,0) \Vert} \} = \mathbb{R}^2.$$
However,
$$ f_c(\mathbb{R}^2)=f_c([0, 2\pi c]\times [0, 2\pi c]) = [-1,1]\times [-1,1],$$
which is not open.
