Part 1 - Determinant of X$X$
As a partial result, it is possible to show that $|\det X| < 1$$\lvert\det X\rvert < 1$ for sufficiently small $\epsilon$. In fact, because $B$ and $C$ commute, then thanks to a known property of block matrices, we have that
\begin{equation} \det\begin{bmatrix} A-\epsilon I & I\\ -\epsilon C & B \end{bmatrix} = \det ((A-\epsilon I)B+\epsilon C). \end{equation} From OP's assumptions on the eigenvalues of $A$ and $B$ we have that $|\det A| < 1$$\lvert\det A\rvert < 1$ and $|\det B| < 1$$\lvert\det B\rvert < 1$, which implies that \begin{equation} |\det(AB)| = |\det A||\det B| < 1. \end{equation}\begin{equation} \lvert\det(AB)\rvert = \lvert\det A\rvert\lvert\det B\rvert < 1. \end{equation} Hence, by continuity there exists sufficiently small $\epsilon>0$ such that \begin{equation} |\det ((A-\epsilon I)B+\epsilon C)| < 1. \end{equation}\begin{equation} \lvert\det ((A-\epsilon I)B+\epsilon C)\rvert < 1. \end{equation}
Part 2 - All eigenvalues except one
Let $0<m<1$ be the maximum modulus of all eigenvalues of $A$ (excluding the eigenvalue $1$) and $B$. The eigenvalues of $A$ and $B$ fulfil \begin{equation} \det((A -\lambda I)(B-\lambda I)) = 0 \end{equation} and, except the eigenvalue $1$ of $A$, their distance from the boundary of the unit disk is at least $1-m$. Then by continuity, for sufficiently small $\epsilon > 0$, all least all solutions except one of the perturbed equation \begin{equation} \det((A-\epsilon I -\lambda I)(B-\lambda I) + \epsilon C) = 0 \end{equation} are in the unit disk, but the solutions of such equation are the eigenvalues of $X$.
Part 3 - The last eigenvalue when $A$ and $B$ are symmetric
Let $v_A$ be the eigenvector corresponding to the single eigenvalue $1$ of $A$. Then $\begin{bmatrix}v_A\\ 0\end{bmatrix}$ is the eigenvector corresponding to the (single) eigenvalue $1$ of the matrix \begin{equation} \begin{bmatrix} A & I\\ 0 & B \end{bmatrix}. \end{equation} Hence, for sufficiently small $\epsilon > 0$, the perturbed block matrix $X$ has a (still single, hence real) eigenvalue $\lambda$ that tends to $1$ as $\epsilon \rightarrow 0$. Let $\begin{bmatrix}v\\ w\end{bmatrix}$ be the corresponding eigenvector. From the bottom blocks of X we have \begin{equation} w = -\epsilon(\lambda I - B)^{-1}Cv. \end{equation} Because $B$ and $C$ commute, so do $(\lambda I - B)^{-1}$ and $C$, which implies that the spectrum of $(\lambda I - B)^{-1}C$ is, up to some ordering, the product of the spectra (see up to some ordering, the product of the spectra@user91684's answer to Eigenvalues of the product of two diagonalizable commuting matrices.) of $(\lambda I - B)^{-1}$ and $C$. Because (i) $\lambda$ is arbitrarily close to 1, (ii) the eigenvalues of $B$ are in ]-1+m,1-m[$\mathopen]-1+m,1-m\mathclose[$ and (iii) $C$ is positive semidefinite, then the eigenvalues $\lambda_1,\dots,\lambda_n$$\lambda_1,\dotsc,\lambda_n$ of $-\epsilon(\lambda I - B)^{-1}C$ are nonpositive and let $v_1,\dots, v_n$$v_1,\dotsc, v_n$ be the respective eigenvectors (which can be chosen orthonormal since $-\epsilon(\lambda I - B)^{-1}C$ is symmetric). Then we can write \begin{equation} v = \sum_{i=1}^n \xi_i v_i;\\ w = \sum_{i=1}^n \lambda_i\xi_i v_i. \end{equation} It follows that \begin{equation} \tag{A0} \langle w,v\rangle = \sum_{i=1}^n \lambda_i\|\xi_i v_i\|^2 \leq 0. \end{equation}\begin{equation} \tag{A0}\label{A0} \langle w,v\rangle = \sum_{i=1}^n \lambda_i\lVert\xi_i v_i\rVert^2 \leq 0. \end{equation} From the top blocks of $X$ we have \begin{equation} \tag{A1} (A-\epsilon I)v + w = \lambda v. \end{equation}\begin{equation} \tag{A1}\label{A1} (A-\epsilon I)v + w = \lambda v. \end{equation} By testing (\eqref{A1)} against $v$, using (\eqref{A0)} and the symmetry of $A$ we have \begin{align} \tag{A2} \lambda\|v\|^2 = \langle w,v\rangle + \langle (A - \epsilon I)v,v\rangle &\leq \langle (A - \epsilon I)v,v\rangle \leq \rho(A - \epsilon I)\|v\|^2\\ &\leq (1-\epsilon) \|v\|^2. \end{align}\begin{align} \tag{A2}\label{A2} \lambda\lVert v\rVert^2 = \langle w,v\rangle + \langle (A - \epsilon I)v,v\rangle &\leq \langle (A - \epsilon I)v,v\rangle \leq \rho(A - \epsilon I)\lVert v\rVert^2\\ &\leq (1-\epsilon) \lVert v\rVert^2. \end{align} Because $v_A \neq 0$, for sufficiently small $\epsilon$, by continuity, also $\|v\| \neq 0$$\lVert v\rVert \neq 0$, so we cancel $\|v\|$$\lVert v\rVert$ on both sides of (\eqref{A2)}, which implies that \begin{equation} \lambda \leq 1-\epsilon \end{equation} when $\epsilon$ is sufficiently small.
