How do we prove that the number of sign changes in the sequence of the determinants of the upper-left matrices of a symmetric matrix $A$ corresponds to the number of positive and negative eigenvalues of $A$?
Progress
I know that it is true for symmetric matrices that if all the upper left determinants are positive, then all the eigenvalues are positive. This is usually stated in texts when positive definite matrices are introduced.
This follows from the min-max principle. We can proceed by induction on the size of $A$. We definitely want to assume that $0$ is not an eigenvalue (otherwise the diagonal matrix starting with a $0$ would defeat us), and then I can also assume that none of the determinants is zero, by slightly perturbing $A$ if necessary.
Note that if the top left $(n-1)\times (n-1)$ block of $A$ (let's call it $A_{n-1}$) has $N_+$ positive and $N_-$ negative eigenvalues, then $A$ itself has at least $N_+$ positive and at least $N_-$ negative eigenvalues. This follows from min-max, because we have subspaces of these dimensions on which the quadratic form $\langle x, Ax \rangle$ is strictly positive and negative, respectively.
Let me now discuss one typical case: $\det A_{n-1}>0$ and $\det A>0$ as well. Then $N_-$ (for $A_{n-1}$) is even, but $A$ must also have an even number of negative eigenvalues, so $N_-$ cannot change when we pass from $A_{n-1}$ to $A$, and thus $N_+$ increases by one.
The other cases are completely analogous.