Can someone please look over my proof below as to why $\det(X+iY) \det(X-iY)>0$ for real matrices $X,Y$, such that $\det(X+iY)$, $ \det(X-iY)$ not both the zero, and tell me if it's correct ?
My original idea was that $\det(X-iY)=\det(\overline{X+iY})=\overline{\det(X+iY)}$ (where the overline, when applied to a matrix, denotes pointwise conjugation of the elements of the matrix), so $\det(X+iY) \det(X-iY)= \det(X+iY) \overline{\det(X+iY)}$, which is $>0$, since for any nonzero complex number $w$, we have that $w\overline{w}>0$.
Yes, your proof is good. The determinant of an $n\times n$ matrix can be written $$ \sum_{\sigma\in S_n} (\operatorname{sgn}\sigma) a_{1,\sigma(1)}a_{2,\sigma(2)}\dots a_{n,\sigma(n)} $$ and conjugating each coefficient gives the conjugate.
Another way for seeing this is to write the matrix as $$ A=PLU $$ where $P$ is a permutation matrix (which has real coefficients), $L$ is lower triangular and invertible, and $U$ is upper triangular (with either $1$ or $0$ along the diagonal); thus $\det P=\det\bar{P}=\pm1$ and $$ \det U=\det\bar{U}=\begin{cases} 0 & \text{if $A$ is not invertible}\\ 1 & \text{if $A$ is invertible} \end{cases} $$ So, in case $A$ is invertible, $\det A=\det L$ and $\det\bar{A}=\det\bar{L}$, but the determinant of a triangular matrix is the product of the coefficients on the diagonal. Thus $\det\bar{L}=\overline{\det L}$ and $\det \bar{A}=\overline{\det A}$ follows.