A simple eigenvalue is stable. We say that an eigenvalue $\lambda$ of a square matrix $A$ is simple if $\det(A-\lambda I)=0$ but 
$$\displaystyle\frac{d}{d\lambda} \det(A-\lambda I)\not =0 $$

Prove that there exist a $r>0$, $c>0$ such that for every $A'$ square matrix such that $\lVert A-A'\rVert<r$ has a simple eigenvalue $\lambda'$ such that
  $$\lvert\lambda-\lambda'\rvert\leq C \lVert A-A'\rVert$$

To prove this I tried to prove that the function: 
$$f(\lambda)=\lambda-k^{-1}\det(I-\lambda A) $$
where $k=\frac{d}{d\lambda} \det(A'-\lambda I)$ is a contraction in a suitable neighborhood of $\lambda_0$ and a suitable neighborhood of $A$ and $A'$ belongs to this neighborhood. 
But I am stuck making this idea concrete. Any advice?
($\lVert A\rVert$ is norm of $A$ as an operator $\sup_{\lVert x\rVert=1}\lvert Ax \rvert$)
 A: My analysis is rusty, but I think the below arguments work.
For all $B\in Mn(\mathbb{R})$ using the operator norm, and for all $s\in \mathbb{C}$, define $$f(B,s) = \det(B-sI).$$ Then $f$ is a continuously differentiable (with respect to the Frechet derivative) function defined on all of $Mn(\mathbb{R})\times \mathbb{C}$. Furthermore, we have that $$f(A,\lambda) = 0\mbox{ and }\frac{\partial}{\partial s} f(B,s)|_{(A,\lambda)} =a\neq 0,$$
where $Ax=\lambda x$, and $\lambda$ is a simple eigenvalue of $A$. Hence, $s\mapsto f(B,s)|_{(A,\lambda)}\cdot(0,s)=(0,as)$ is a Banach isomorphism from $\mathbb{C}$ onto $\mathbb{C}$. Therefore, by the implicit function theorem (defined for Banach spaces) there exists an open neighborhood $U_0$ of $A$ in $Mn(\mathbb{R})$ such that for all open balls $U\subset U_0$ containing $A$ there is a unique continuously differentiable function
$u: Mn(\mathbb{R}) \to \mathbb{C}$ such that $u(A) = \lambda$, $(B, u(B)) \in Mn(\mathbb{R})\times \mathbb{C}$  and $f(B, u(B)) = 0$ for all $B\in U$. Let $U$ be such a ball with radius $\epsilon$. Choose $\epsilon>r>0$ such that for all $B$ where $||A-B||\leq r$ we have that $B\in U$. Then there exists a $C$ bounding the derivative $Du(B)$ with respect to the operator norm and we have for $\mu=u(B)$ that $$|\lambda - \mu| \leq C ||A-B||$$ as desired.
