Since $A$ is nonsingular, consider the following block factorization of $C$:
$$
C=\pmatrix{A&B^T\\B&0}
=
\pmatrix{I&0\\BA^{-1}&I}\pmatrix{A&0\\0&S}\pmatrix{I&A^{-1}B^T\\0&I},
$$
where $S:=-BA^{-1}B^T$. Since the triangular blocks are nonsingular, the matrix $C$ is nonsingular iff the Schur complement matrix $S$ is nonsingular.
Now if $A$ is SPD, it is easy to see that $S$ is SPD as well. First, the definiteness of $A$ implies that $A^{-1}$ is SPD. For a nonzero $x$, $B^Tx\neq 0$ since has $B$ has full row rank, and $$x^T(BA^{-1}B^T)x=(B^Tx)^TA^{-1}(B^Tx)>0.$$
Another way to see that $C$ is nonsingular if $A$ is SPD and $B$ has full row rank is as follows. Assume that $Cz=0$ for some nonzero $z=(x^T,y^T)^T$. Hence
$$\tag{1}
Ax+B^Ty=0, \quad Bx=0.
$$
None of the block components can be zero. If $x=0$ and $y\neq 0$ then $B^Ty=0$ which is impossible since $B$ has full row rank. If $x\neq 0$ and $y=0$ then $Ax=0$ which is impossible since $A$ is SPD. Hence both $x\neq 0$ and $y\neq 0$. Multiply the first equation in (1) with $x^T$ and the second with $y^T$ to get
$$
x^TAx+x^TB^Ty=0, \quad y^TBx=0.
$$
Since $x^TB^Ty=y^TBx=0$, we have $x^TAx=0$, which gives again a contradiction.
It is not sufficient that $A$ is nonsingular and $B$ of full rank for $S$ being nonsingular. Consider,
$$
A=\pmatrix{1 & 0 \\ 0 & -1},
\quad B=(1,1).
$$
It is easy to verify that $C$ is singular (actually $S=0$).
Best Answer
Another fact that wasn't pointed out is that since the determinant is a product of the eigenvalues and both PD and ND matrices have either all strictly positive or all strictly negative eigenvalues you can deduce the determinant is non zero so the matrix is invertible (i.e. has full rank).