Condition of the eigenvalue problem [Ciarlet, 2.3-1] I know this result:

Let $A$ a diagonalisable matrix, $P$ a matrix such that
  $$P^{-1}AP\ =\ \mbox{diag}(\lambda_i)\ =\ D,$$
  and $\|\cdot\|$ a matrix norm satisfying
  $$\|\mbox{diag}(d_i)\|\ =\ \max_i|d_i|$$
  for every diagonal matrix. Then, for every matrix $\delta A$,
  $$\mbox{sp}(A+\delta A)\ =\ \{\mu : (A+\delta A)x=\mu x, \mbox{ for some } x\in\mathbb{C}^n\}\ \subset\ \bigcup_{i=1}^ND_i,$$
  where
  $$D_i\ =\ \{z\in\mathbb{C}:|z-\lambda_i|\leq\mbox{cond}(P)\|\delta A\|\}.$$

Now, I want to prove that: If there exists an integer $m$ satisfying $1\leq m\leq n$ such that the union $\bigcup\limits_{i=1}^mD_i$ of the $m$ disk $D_i$ is disjoint form the union $\bigcup\limits_{i=m+1}^nD_i$ of the remaining $n-m$ disks (it is always possible to assum that it is the first $m$ and last $n-m$ disks which have this property), then the union $\bigcup\limits_{i=1}^mD_i$ contains exactly $m$ eigenvalues of the matrix $A + \delta A$.

Please, somebody have an idea to solve it? Thanks in advance.
 A: Define $\hat{A}(\alpha) = A+\alpha\delta A$ and note that $\mathrm{sp}(\hat{A}(\alpha))\subseteq\bigcup_{i=1}^N D_i$ for $\alpha\in[0,1]$.
Recall that the eigenvalues of a matrix depend continuously on the coefficients of said matrix.
Therefore, we have continuous paths $\lambda_i\colon[0,1]\to\bigcup_{j=1}^N D_j$ such that $\mathrm{sp}(\hat{A}(\alpha)) = \{\lambda_i(\alpha)|i\in\{1,\ldots,N\}\}$.
Without loss of generality we may label the $\lambda_i$ such that $\lambda_i(0)\in D_i$.
Since $\bigcup_{i=1}^m D_i$ is disjoint from $\bigcup_{i=m+1}^N D_i$ and since both are closed, no path can start in the former and end in the latter. For $i\in\{1,\ldots,m\}$, $\lambda_i$ is a path starting in the former and hence ending in the former as well.
Topological intermezzo:
Suppose $A$ and $B$ are closed sets and suppose that $p\colon[0,1]\to A\cup B$ is a continuous path such that $p(0)\in A$ and $p(1)\in B$.
By continuity of $p$, $p^{-1}(A)\subseteq [0,1]$ is closed as well as $p^{-1}(B)\subseteq [0,1]$.
Furthermore, $[0,1] = p^{-1}(A\cup B) = p^{-1}(A)\cup p^{-1}(B)$, thus since both are closed $\inf(p^{-1}(B))\in p^{-1}(A)\cap p^{-1}(B)$.
It follows that $p(\inf(p^{-1}(B)))\in A\cap B$ hence $A$ and $B$ are not disjoint.
By contraposition, if $A$ and $B$ are closed and disjoint, no continuous path $p$ can exists such that $p(0)\in A$ and $p(1)\in B$.
Thus, we conclude $\lambda_1(1),\ldots,\lambda_m(1)\in\bigcup_{i=1}^m D_i$, where $\lambda_1(1),\ldots,\lambda_m(1)\in\mathrm{sp}(A+\delta A)$.
By the same argument $\lambda_{m+1}(1),\ldots,\lambda_{N}(1)\in\bigcup_{i=m+1}^N D_i$, where $\lambda_{m+1}(1),\ldots,\lambda_N(1)\in\mathrm{sp}(A+\delta A)$ proving your conjecture.
