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 $\epsilon$.