Here is a proof of $\det M\geq 0$ for $M\in \mathfrak{sp}(n)$ based on the lemma that *every complex matrix is <A HREF="https://en.wikipedia.org/wiki/Matrix_consimilarity">consimilar</A> to a real matrix.*

*Acknowledgment:* In what follows I was helped by feedback I received at <A HREF="https://math.stackexchange.com/q/4039880/87355">MSE</A>.

By construction, the $2n\times 2n$ complex matrix $M\in \mathfrak{sp}(n)$ is <A HREF="https://en.wikipedia.org/wiki/Skew-Hermitian_matrix">skew-Hermitian</A> and <A HREF="https://en.wikipedia.org/wiki/Hamiltonian_matrix">Hamiltonian</A>, which means that it has the $n\times n$ block decomposition
$$M=\begin{pmatrix}
A&B\\ C&-A^T\end{pmatrix},\;\;\text{with}\;\;A=-A^\ast,\;\;B=B^T=-C^\ast=-\bar{C}.$$
Here $M^T$ denotes the transpose, $\bar{M}$ the complex conjugate, and $M^\ast$ the conjugate transpose.  

By continuity of the determinant it is sufficient to consider $\det A\neq 0$. Then <A HREF="https://en.wikipedia.org/wiki/Schur_complement">Schur's determinant identity</A> gives

$$\det M=\det(-AA^T-ACA^{-1}B)=\det(A\bar{A}+A\bar{B}A^{-1}B)$$
$$\qquad=\det(A\bar{A})\det(1+\bar{A}^{-1}\bar{B}A^{-1}B)$$
$$\qquad=|\det A|^2\det(1+\bar{X}X),\;\;\text{with}\;\;X=A^{-1}B.$$

Now I apply the consimilarity lemma, to write $X=SR\bar{S}^{-1}$ with $R$ a real matrix. This gives
$$\det M=|\det A|^2\det(1+\bar{S}R^2\bar{S}^{-1})=|\det A|^2\det(1+R^2)$$
$$\qquad=|\det A|^2\det(1+iR)\det(1-iR)$$
$$\quad=|\det A|^2|\det(1+iR)|^2\geq 0.$$