Sylvester's criterion says that an $n\times n$ Hermitian matrix $A$ is positive definite if and only if all its leading principal minors of are positive. If one knows that fact that every Hermitian matrix has an orthogonal eigenbasis, one can prove Sylvester's criterion easily.
The forward implication is obvious. If $A$ is positive definite, so are all its leading principal submatrices. Their spectra are hence positive. Thus the leading principal minors are positive, because each of them is a product of the eigenvalues of the submatrix.
To prove the backward implication, we use mathematical induction on $n$. The base case $n=1$ is trivial. Suppose $n\ge2$ and all leading principal minors of $A$ are positive. In particular, $\det(A)>0$. If $A$ is not positive definite, it must possess at least two negative eigenvalues. As $A$ is Hermitian, there exist two mutually orthogonal eigenvectors $x$ and $y$ corresponding to two of these negative eigenvalues. Let $u=\alpha x+\beta y\ne0$ be a linear combination of $x$ and $y$ such that the last entry of $u$ is zero. Then
u^\ast Au=|\alpha|^2x^\ast Ax+|\beta|^2y^\ast Ay<0.
Hence the leading $(n-1)\times(n-1)$ principal submatrix of $A$ is not positive definite. By induction assumption, this is impossible. Hence $A$ must be positive definite.