Part 1: Eigenvalues lie in the union of discs. Let $\lambda$ be an eigenvalue with eigenvector $x = (x_1, \ldots, x_n)^T$. Let $x_i$ be the entry with maximum absolute value: $|x_i| = \|x\|_\infty > 0$.

From $Ax = \lambda x$, the $i$-th equation gives: $\sum_{j=1}^n a_{ij} x_j = \lambda x_i$.

Isolating the diagonal term: $(\lambda - a_{ii}) x_i = \sum_{j \neq i} a_{ij} x_j$.

Taking absolute values: $|\lambda - a_{ii}| |x_i| \leq \sum_{j \neq i} |a_{ij}| |x_j| \leq \sum_{j \neq i} |a_{ij}| |x_i| = R_i |x_i|$.

Dividing by $|x_i| > 0$: $|\lambda - a_{ii}| \leq R_i$, so $\lambda \in D_i$.

Part 2: Counting eigenvalues in connected components. Consider the family of matrices $A(t) = D + t(A - D)$ for $t \in [0, 1]$, where $D = \mathrm{diag}(a_{11}, \ldots, a_{nn})$. At $t = 0$, the eigenvalues are $a_{11}, \ldots, a_{nn}$ with the $i$-th eigenvalue in the degenerate disc $\{a_{ii}\}$. At $t = 1$, we recover $A$.

The Gershgorin discs for $A(t)$ are $D_i(t) = \{z : |z - a_{ii}| \leq tR_i\}$. As $t$ increases from 0 to 1, the discs grow continuously, and the eigenvalues (as continuous functions of $t$) move continuously within the union of discs.

If $m$ discs form a connected component $\Omega(t)$ that remains isolated from the other $n - m$ discs for all $t \in [0, 1]$ (i.e., $\Omega(t) \cap D_j(t) = \emptyset$ for $j$ outside the component), then no eigenvalue can enter or leave $\Omega(t)$ as $t$ varies. At $t = 0$, exactly $m$ eigenvalues (the $m$ diagonal entries) lie in $\Omega(0)$. By continuity, exactly $m$ eigenvalues lie in $\Omega(1)$. $\square$