Clearly all eigenvalues apart from the eigenvalue $\lambda(\epsilon)$ with $\lambda(0) = 1$ stay in the open unit disk for $\epsilon$ sufficiently small. To see what happens to the last eigenvalue, use eigenvalue first-order perturbation theory, for instance Theorem 1 in https://arxiv.org/abs/1903.00785 : if $v$ and $w^*$ are the right and left eigenvalue associated to $\lambda(0)$, then the right and left eigenvectors of the $2\times 2$ work matrix are $\begin{bmatrix}v\\0\end{bmatrix}$ and $\begin{bmatrix}w^* & -w^*B\end{bmatrix}$ and multiply them by $\begin{bmatrix}I & 0\\-C & 0\end{bmatrix}$. $$ \frac{d\lambda(\epsilon)}{d\epsilon} = .... $$
As the derivative is negative, for a sufficiently small $\epsilon > 0$ we have that $\lambda(\epsilon)$ is in the unit disk.