I assume that the OP meant
$$\begin{bmatrix} \mathrm A \,\, & \mathrm X\\ \mathrm X^{\top} & \mathrm B\end{bmatrix} \succeq \mathrm O$$
where $\mathrm X = \mathrm A^{\frac 12} \mathrm C \, \mathrm B^{\frac 12}$ and $\mathrm A, \mathrm B \succeq \mathrm O$. Hence,
$$\begin{bmatrix} \mathrm A^{\frac 12} \mathrm A^{\frac 12} & \mathrm A^{\frac 12} \mathrm C \, \mathrm B^{\frac 12}\\ \mathrm B^{\frac 12} \mathrm C^{\top} \mathrm A^{\frac 12} & \mathrm B^{\frac 12} \mathrm B^{\frac 12}\end{bmatrix} = \begin{bmatrix} \mathrm A^{\frac 12} & \\ & \mathrm B^{\frac 12}\end{bmatrix} \begin{bmatrix} \mathrm I & \mathrm C\\ \mathrm C^{\top} & \mathrm I\end{bmatrix} \begin{bmatrix} \mathrm A^{\frac 12} & \\ & \mathrm B^{\frac 12}\end{bmatrix} \succeq \mathrm O$$
which holds if
$$\begin{bmatrix} \mathrm I & \mathrm C\\ \mathrm C^{\top} & \mathrm I\end{bmatrix} \succeq \mathrm O$$
Using the Schur complement, the LMI above can be rewritten in the form
$$\mathrm I - \mathrm C^{\top} \mathrm C \succeq \mathrm O$$
Let us consider the special case where $\rm C$ is symmetric and, thus, its eigenvalues are real. Let its spectral decomposition be $\rm C = Q \Lambda Q^{\top}$. Hence,
$$\mathrm I - \mathrm C^{\top} \mathrm C = \mathrm I - \mathrm C^2 = \mathrm Q \, \left( \mathrm I - \Lambda^2 \right) \, \mathrm Q^{\top} \succeq \mathrm O$$
which holds if
$$\mathrm I - \Lambda^2 \succeq \mathrm O$$
Since $\rm C$ is a contraction matrix, its spectral radius is less than $1$. Thus, $\mathrm I - \Lambda^2 \succeq \mathrm O$ does indeed hold. We conclude that
$$\rho (\mathrm C) < 1 \text{ and } \mathrm C = \mathrm C^{\top} \implies \begin{bmatrix} \mathrm A \,\, & \mathrm X\\ \mathrm X^{\top} & \mathrm B\end{bmatrix} \succeq \mathrm O$$
The general (non-symmetric) case is left as an exercise for the reader.