Consider the set $\{(x,t)\in\mathbb R^n\times\mathbb R\mid t>\|x\|\}$ ad the matrices:
$$ A=(t^2-x^Tx)I+2xx^T,\quad B=-2tx,\quad C=t^2+x^Tx. $$
Clearly $C\succ 0$ (it is a positive scalar). I need to show that $A-BC^{-1}B^T\succ 0$.
My attempt at a solution has been to substitute the values for $A$, $B$ and $C$ and massage the expression to obtain:
$$ A-BC^{-1}B^T=\underbrace{(t^2-\|x\|^2)}_{>0}\left(I-\frac{2}{t^2+\|x\|^2}xx^T\right). $$
However it does not look to me like $I-\frac{2}{t^2+\|x\|^2}xx^T$ is positive definite... $I$ certainly is, but $-\frac{2}{t^2+\|x\|^2}xx^T$ is a rank 1 matrix with the eigenvalues $\{-\frac{2}{t^2+\|x\|^2},0\}$ so it is negative semidefinite.
We can complete the proof by using the definition of positive definite, as per max_zorn's suggestion. Consider $y\in\mathbb R^n,y\ne 0$. We have:
$$ y^T\left(I-\frac{2}{t^2+\|x\|^2}xx^T\right)y=\frac{t^2\|y\|^2+\|x\|^2\|y\|^2-2\|x^Ty\|^2}{t^2+\|x\|^2}\ge \frac{t^2\|y\|^2+\|x\|^2\|y\|^2-2\|x\|^2\|y\|^2}{t^2+\|x\|^2}>0 $$
where the first inequality is due to Cauchy-Schwarz inequality.