In control theory, this kind of inequalities is ubiquitous and handled via going back and forth using Schur complements. For completeness, here is the non-strict version of Schur complement formula, it is an overkill but the question is a particular special case, so here it goes:
Formula: Let $Q,R$ be symmetric matrices. Then following are equivalent:
$ \begin{pmatrix} Q &S\\ S^T &R \end{pmatrix} \succeq 0 $
$\begin{align} R &\succeq 0\\ Q -SR^\dagger S^T &\succeq 0\\ S(I-RR^\dagger) &= 0 \end{align} $ where $R^\dagger$ is the Pseudoinverse of $R$.
Now if renaming your hypothesis involving the matrix difference as $ M_1 - M_2 := \begin{pmatrix} A_1 &B_1\\ B_1^T &C_1 \end{pmatrix} - \begin{pmatrix} A_2 &B_2\\ B_2^T &C_2 \end{pmatrix}\succeq 0$
we can reformulate the hypothesis as the following via the second item of the formula ($Q=M_1,S=I,R=M_2^{-1}$): $\begin{align} M_2^{-1} &\succeq 0 \quad \text{by definition}\\ M_1 - M_2 &\succeq 0\\ I (I-M_2M_2^{-1}) &= 0 \end{align} $ hence we have $ \begin{pmatrix} M_1 &I\\I&M_2^{-1} \end{pmatrix}\succeq 0 $ Also, from the inverse of a matrix formula, we have $ M_2^{-1} = \begin{pmatrix} (A_2 - B_2C_2^{-1}B_2^T)^{-1} &\star\\ \star &\star \end{pmatrix} $ As user1551 showed, you can bring the matrix $M_1$ in the form of the following by a congruence transformation and some reshuffling: $\begin{pmatrix} M_1 &I\\I&M_2^{-1} \end{pmatrix} \leadsto \left( \begin{array}{cc|cc} (A_1 - B_1C_1^{-1}B_1^T) &I &0 &0\\ I &(A_2 - B_2C_2^{-1}B_2^T)^{-1} &0 &\star\\ \hline 0 &0 &C_1 &I\\ 0&\star&I&\star \end{array}\right) \succeq 0 $ The $(1,1)$ block matrix has the desired result if we apply the nonstrict Schur complement formula again but in the reverse direction.
Last minute addition: Now that I look at it, it is not as good as I thought it to be initially but I did not want to waste the whole thing so I hope it helps an $\epsilon$.