If
$$A=\begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix}$$
is a partition of $A$ such that $A_{11}$ and $A_{22}$ are $r \times r$ and $(n − r) \times (n − r)$ matrices, respectively, then
$$\det(A) \leq \det(A_{11}) \cdot \det(A_{22})$$
with equality holding if and only if $A$ is block diagonal,
Hint: Use $\det(A+B)^{\frac{1}{n}} \geq \det(A)^{\frac{1}{n}} + \det(B)^{\frac{1}{n}}$ with
$$B=\begin{bmatrix} A_{11} & -A_{12} \\ -A_{21} & A_{22} \\ \end{bmatrix}$$
Why is $B$ positive definite?
It's not true. Consider, e.g. $n=2r$ and $A=\pmatrix{0&-I_r\\ I_r&0}$.
Edit. If $A$ is positive definite, the inequality in question is true. To prove it, you may apply the other inequality --- known as Minkowski's inequality --- in the hint. Minkowski's inequality requires that both $A$ and $B$ are positive definite. Let $D=D^T=\operatorname{diag}(I_r,-I_{n-r})$ and $B=D^TAD$. Then $A+B=2\,\operatorname{diag}(A_{11},A_{22})$. As the hint asks, you have to explain why $B$ is positive definite (actually you need to explain why $A+B$ is positive definite as well). Apply Minkowski's inequality, the result follows immediately.