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
Is a full rank square matrix necessarily a positive definite matrix?
No, as pointed out in the comments.
Is a full rank controllability Grammian necessarily positive definite?
Yes. To see this note that the controllability Grammian is defined to be
$$ W(t)\triangleq\int_{0}^{t}e^{A\tau}BB^{T}e^{A^{T}\tau}d\tau. $$
Now suppose that $W(t)$ is full rank. It follows that $\ker(W) = \{0\}$, and thus, for all $x\in \mathbb{R}^n\setminus\{0\}$, $x^TW(t)x \neq 0$. Furthermore
\begin{align} x^TW(t)x &= \int_{0}^{t}x^Te^{A\tau}BB^{T}e^{A^{T}\tau}xd\tau\\ % &= \int_{0}^{t}(B^{T}e^{A^{T}\tau}x)^T(B^{T}e^{A^{T}\tau}x)d\tau\\ % &= \int_{0}^{t}\|B^{T}e^{A^{T}\tau}x\|^2d\tau.\\ \end{align}
Since the norm of a vector is always positive or zero it follows that the integral is positive or zero. Since the quadratic form cannot be zero, by virtue of the fact that $W(t)$ is full rank, it follows that the integral must be positive, and thus $x^TW(t)x > 0$, which is the condition that $W(t)$ be positive definite.