I am trying to follow some notes that state $$ M= \begin{bmatrix} A&B^T\\B&C \end{bmatrix} \succeq 0 \Longleftrightarrow M'= \begin{bmatrix} A&-B^T\\-B&C \end{bmatrix} \succeq 0$$ and I want to prove this to myself (just the $\Rightarrow$ direction because it's trivial to go the other way once one direction is proven).
Clearly, $ \begin{bmatrix} A&B^T\\B&C \end{bmatrix} \succeq 0\Longrightarrow A,C\succeq0 $ by computing $x^TMx$ (for $x=\begin{bmatrix}v\\0\end{bmatrix}$ and $\begin{bmatrix}0\\v\end{bmatrix}$) which is $\geq0\ \ \forall\ v$ by definition.
Then for a general $x=\begin{bmatrix}x_1\\x_2\end{bmatrix}$,
$$ \begin{bmatrix} x_1^T&x_2^T \end{bmatrix} M \begin{bmatrix} x_1\\x_2 \end{bmatrix} = \underbrace{x_1^TAx_1}_{\geq0} + \underbrace{x_2^TCx_2}_{\geq0} +2x_2^TBx_1\geq2x_2^TBx_1 $$
and
$$ \begin{bmatrix} x_1^T&x_2^T \end{bmatrix} M' \begin{bmatrix} x_1\\x_2 \end{bmatrix} =x_1^TAx_1+x_2^TCx_2-2x_2^TBx_1\geq-2x_2^TBx_1. $$
Then I get stuck.
Just apply your assumption on $M$ to the vector $[-x_1^T,x_2^T]$.