It is an exercise in Peter Lax's book Linear Algebra that if all the Gershgorin disks
$$D_i := \{z\in \mathbb{C} : |a_{ii} - z| \leq \sum_{i \neq j} |a_{ij}|\}$$
are disjoint, then each disk must contain exactly one eigenvalue of the matrix $A = (a_{ij})$.
What I've tried so far: The proof of Gershgorin's theorem involves selecting the largest-modulus entry $v_i$ of an eigenvector $v$, and then showing that the corresponding eigenvalue $\lambda$ must be in $D_i$.
It would suffice if we could show that for each $i \in \{1,2,\dotsc, n\}$ there was an eigenvector $v$ such that $\max_{j}|v_j| = i$. But I'm not sure how we could show this...
Let $D$ be the diagonal part of $A$ and $M=A-D$ so that $A=D+M$. Let $\lambda_i(s)$ be the $i$th eigenvalue, counting multiplicities, of $D+sM$.
When $s=1$, we have our original case. When $s=0$, we have a diagonal matrix. As $s$ increases from $0$ to $1$, we are blowing up the radii of the disks given by Gershgorin's theorem.
Keeping this in mind, and noting that $\lambda_i(s)$ is continuous (implicit function theorem), consider that one of the disjoint disks $D_i$ does not have an eigenvalue. Then think about the situation as $s \to 0$. You can work out the contradiction--it's great to think about geometrically.