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 $2\times 2$ matrix in the text are $\begin{bmatrix}v\\0\end{bmatrix}$ and $\begin{bmatrix}w^* & w^*(I-B)^{-1}\end{bmatrix}$ and one gets $$ \frac{d\lambda(\epsilon)}{d\epsilon} = \frac{1}{w^*v} \begin{bmatrix}w^* & w^*(I-B)^{-1}\end{bmatrix}\begin{bmatrix}-I & 0\\-C & 0\end{bmatrix} \begin{bmatrix}v\\0\end{bmatrix} = -1 - \frac{1}{w^*v} w^*(I-B)^{-1}Cv. $$ If the derivative is negative, for a sufficiently small $\epsilon > 0$ we have that $\lambda(\epsilon)$ is in the unit disk.
If $B$ and $C$ commute you can go on and simplify the last expression, but in general it should be possible to construct cases in which the derivative is positive.