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 $z^*$ 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 $\begin{bmatrix}z^* & z^*(I-B)^{-1}\end{bmatrix}$ and one gets $$ \frac{d\lambda(\epsilon)}{d\epsilon} = \frac{1}{z^*v} \begin{bmatrix}z^* & z^*(I-B)^{-1}\end{bmatrix}\begin{bmatrix}-I & 0\\-C & 0\end{bmatrix} \begin{bmatrix}v\\0\end{bmatrix} = -1 - \frac{1}{z^*v} z^*(I-B)^{-1}Cv. $$ If this derivative is negative, for a sufficiently small $\epsilon > 0$ we have that $\lambda(\epsilon)$ is in the unit disk.
If $B$ and $C$ commute and $B$ is symmetric you can go on and simplify the last expression as in the other answer, but in general it should be possible to construct cases in which the derivative is positive and hence the result does not hold. In the end, $v$ and $z$ can be chosen arbitrarily and independently of $B$ and $C$.