Let $H$ be some Hermitian $2N\times 2N$ matrix with complex entries. Write it as a block matrix via $$ H=\begin{bmatrix}A & B\\B^\ast & C\end{bmatrix} $$ where $A,B,C$ are $N\times N$ matrices with complex entries and $A,C$ are furthermore Hermitian.
We know easily that $H$ has real eigenvalues and furthermore $$ \|(H-zI)^{-1}\|\leq\frac{1}{|\Im\{z\}|} \qquad(z\in\mathbb{C})$$ where $\|\cdot\|$ is the matrix operator norm.
Can this knowledge somehow be used to assert the existence of the Schur complement for $$H-zI\,?$$
What I mean is as follows: We write $$ H-zI = \begin{bmatrix}A-zI & B\\B^\ast & C-zI\end{bmatrix} $$ and the Schur formula says that if the Schur complement $$S(z) \equiv A-zId-B(C-zI)^{-1}B^\ast $$ is invertible and so is $C-zI$ (which is trivial if $\Im\{z\}\neq 0$ as $C$ is Hermitian) then $$ (H-zI)^{-1} = \begin{bmatrix}S(z)^{-1} & -S(z)^{-1}B(C-zI)^{-1}\\-(C-zI)^{-1}B^\ast S(z)^{-1} & (C-zI)^{-1}+(C-zI)^{-1}B^\ast S(z)^{-1}B(C-zI)^{-1}\end{bmatrix} \,. $$
However, for $z\notin\mathbb{R}$, the Hermitian nature of $H$ somehow can't seem to be immediately used for the invertibility of $S(z)$ since $S(z)$ is not Hermitian when $z\notin\mathbb{R}$. Indeed, we have $$ [(C-zI)^{-1}]^\ast = (C-\bar{z}I)^{-1}\,. $$
So it seems like to conclude that $S(z)$ is invertible off the real axis we must require that $$\|B\| < |\Im\{z\}|\,,$$ i.e., we cannot get all the way down to the real axis. Is there a way around this, to conclude that there is always a Schur-complement arbitrarily near to the real axis?