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 Greenbaum, Li, and Overton - First-order Perturbation Theory for Eigenvalues and Eigenvectors: if $v$ and $z^*$ are the right and left eigenvalue of $A$ associated to $\lambda(0)=1$, then the right and left eigenvectors of the $2\times 2$ matrix $X$ for $\epsilon=0$ are $\begin{bmatrix}v\\0\end{bmatrix}$ and $\begin{bmatrix}z^* & z^*(I-B)^{-1}\end{bmatrix}$ respectively, and plugging these expressions into the theorem 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, then for a sufficiently small $\epsilon > 0$ we have that $\lambda(\epsilon)$ is in the unit disk.
However, it is possible to choose $z,v,B$ such that $z^*(I-B)^{-1}v < 0$, and $C = \alpha I$; then for sufficiently large $\alpha > 0$ the derivative is larger than $0$ and I don't think that the result holds: $X(\epsilon)$ does not have eigenvalues in the unit disk for a sufficiently small $\epsilon > 0$.
For instance, $A = \begin{bmatrix}1 & 8 \\ 0 & 0.9\end{bmatrix}$, $B = \begin{bmatrix}1/4 & -1/4\\ -1/4 & 1/4\end{bmatrix}$, $C=I$ numerically gives me matrices $X$ with eigenvalues larger than $1$ for small $\epsilon$.