$f:U\to\mathbb{R}^n$ is of class $C^1$. If for $a\in U$ ($U$ open), $f'(a)$ is injective, exists ball where $|f(x)-f(y)\ge c|x-y||$ 
$f:U\to\mathbb{R}^n$ is of class $C^1$ in the open $U\subset
 \mathbb{R}^m$. If for some $a\in U$,
  $f'(a):\mathbb{R}^m\to\mathbb{R}^n$ is injective, then there exists
  $\delta>0$ and $c>0$ such that $B = B(a,\delta)\subset U$ and, for any
  $x,y\in B$ we have  $|f(x)-f(y)\ge c|x-y||$. In particular, the
  restriction $f_{|_ B}$ is injective.

Proof given by book:

The function $u\to |f'(a)\cdot u|$ is positive in all the points $u$
  of the unit sphere $S^{m-1}$ (what - 1), which is compact. By the Weiestrass
  theorem, there exists $c>0$ such that $|f'(a)\cdot u| \ge 2c$ for all
  $u\in S^{m-1}$ (what? - 2). By linearity, follow that $|f'(a)\cdot
 v|\ge 2c\cdot |v|$ for all $v\in \mathbb{R}^m$ (what? - 3). For all
  $x\in U$, we write:
$$r(x) = f(x)-f(a) -f'(a)(x-a)$$
Then, for any $x,y\in U$ we have:
$$f(x)-f(y) = f'(a)\cdot (x-y) + r(x)-r(y)$$
By $|u+v|\ge |u|-|v|$ follows:
$$|f(x)-f(y)|\ge |f'(a)\cdot(x-y)|-|r(x)-r(y)|\ge\\ 2c\cdot
 |x-y|-|t(x)-r(y)|$$
Observe that $r$, as defined above, is of class $C^1$, with $r(a) =
> 0$. By the continuity of $r'$, there exists $\delta >0$ such that
  $|x-a|<\delta \implies x \in U $ and $|r'(x)|<c$. The mean value
  inequality applied in $r$ in the convex set $B = B(x,\delta)$ makes
  $x,y\in B \implies |f(x)-f(y)|\ge 2c|x-y|-|x-y|$, that is,
  $$|f(x)-f(y)|\ge c|x-y|$$

what - 1: couldn't it be $0$?
what - 2: if it's already positive, then why do I need Weiertrass theorem? If it's positive then it's always greater than a $c$, I can just make it smaller so it's bigger than $2c$
what - 3: how linearity makes that possible?
Also, which Weiertrass theorem is applied here? There are many.
 A: What 1. The linear map $\mathrm{d}_af$ is injective, hence there is no nonzero vector in its kernel. In particular, for any $u$ of norm $1$, $\mathrm{d}_af.u$ is nonzero.
What 2. The linear map $\mathrm{d}_af$ is continuous, hence $u\mapsto |\mathrm{d}_af.u|$ is continuous and nonzero on a compact set (the sphere). Therefore, it admits a nonzero minimum (I believe this is what is called Weierstrass theorem, a continuous function on a compact set admits both a maximum and a minimum). The point is that the minoration is uniform, namely $c$ does not depend on the point $u$. Knowing What 1. only ensures that for all $u$ there is $c(u)$ such that the required inequality is satisfied. 
I would like to emphasise on this point giving an example. The function $x\mapsto 1/x$ is positive on $]0,+\infty[$ but you cannot find a $c$ such that for all $x>0$, $1/x\geqslant c$. This phenomenon is due to $]0,\infty[$ failing to be compact, you can approach $0$ as close as you want without leaving $]0,+\infty[$.
What 3. If $v$ is a nonzero vector, then $\displaystyle v=|v|\frac{v}{|v|}$, where $\displaystyle\frac{v}{|v|}$ is a unit vector. Hence, using What 2.:
$$\left|\mathrm{d}_af.\frac{v}{|v|}\right|\geqslant 2c.$$
But using the linearity of $\mathrm{d}_af$, one has:
$$\mathrm{d}_af.v=|v|\mathrm{d}_af.\frac{v}{|v|}.$$
More than the linearity, we are using the homogeneity.
This is a common method and a good idea to keep in mind that when dealing with homogeneous problems, it suffices to solve it on the unit sphere which is compact (in the finite-dimensional case). For example, this is used when characterising bounded multilinear map on normed vector spaces. Once again using the same Weierstrass theorem, one derives the continuity of multilinear maps in the finite-dimensional space.
Remark. I am using $\mathrm{d}_af$ for $f'(a)$, I cannot personally bear the notation $f'(a)$.
