Sylvester's criterion and the openness of the positive cone

($\Rightarrow$). If $S>0$ and $x\in\mathbb{R}^k\setminus0$, extend $x$ by zeros to $\tilde x\in\mathbb{R}^n$: $x^TS_kx=\tilde x^TS\tilde x>0$. So each leading block $S_k$ is positive definite, its eigenvalues are positive and $\Delta_k=\det S_k>0$.

($\Leftarrow$). Assume all $\Delta_k>0$. We do Gaussian elimination without row exchanges. The first pivot is $\Delta_1>0$. After eliminating the first column, the lower right block is the Schur complement and is again symmetric; inductively the $k$-th pivot is $d_k=\Delta_k/\Delta_{k-1}$ (with $\Delta_0=1$), since the determinant of the leading $k\times k$ block is the product of the first $k$ pivots and elimination does not change leading minors. All pivots are $>0$, so the elimination never breaks down, and we obtain
$$S=LDL^T,\qquad L\text{ unit lower triangular},\ D=\operatorname{diag}(d_1,\dots,d_n).$$
Therefore for $x\ne0$, with $y=L^Tx\ne0$ ($L$ is invertible),
$$x^TSx=y^TDy=\sum_kd_ky_k^2>0.$$

Openness. The functions $\Delta_k:\mathrm{Sym}_n\to\mathbb{R}$ are polynomials, so continuous, and
$$\mathrm{PD}=\bigcap_{k=1}^n\Delta_k^{-1}\big((0,\infty)\big)$$
is open.

Example. $S=\begin{pmatrix}2&-1\\-1&2\end{pmatrix}$: $\Delta_1=2$, $\Delta_2=3$, so $S>0$ and $S=\begin{pmatrix}1&0\\-\frac12&1\end{pmatrix}\begin{pmatrix}2&0\\0&\frac32\end{pmatrix}\begin{pmatrix}1&-\frac12\\0&1\end{pmatrix}$.

Remark. The criterion needs *leading* minors: $S=\begin{pmatrix}0&0\\0&-1\end{pmatrix}$ has $\Delta_1=0$ and $S_{22}=-1<0$. For positive *semi*definiteness one needs *all* principal minors $\ge0$, not only the leading ones: $\begin{pmatrix}0&0\\0&-1\end{pmatrix}$ has leading minors $0,0$ yet is not semidefinite.