Here is the given question and my work so far:
Question: Let $A$ be an $n \times n$ invertible matrix, and let $u$ and $v$ be two vectors in $\mathbb{R}^n$. Find the necessary and sufficient conditions on $u$ and $v$ in order that the matrix $$ B=\begin{bmatrix} A & u\\[1ex] v^T & 0 \end{bmatrix} $$ be invertible, and give a formula for the inverse when it exists.
Solution: Performing block Gaussian elimination on $B$, specifically $R_2-v^T A^{-1}R_1\longrightarrow R_2$, we obtain
$$ B=\begin{bmatrix} A & u\\[1ex] 0 & -v^TA^{-1}u \end{bmatrix} $$
Next, recall that in order for a matrix to be nonsingular, it is both necessary and sufficient that its determinant be nonzero. Before we proceed, take note of the fact that from $(-v^T A^{-1} u)$ we produce $(1\times n) \cdot (n\times n) \cdot (n\times 1) = 1\times 1$. Thus, the product produces a scalar.
Now, we wish to show that $\det(B)\neq 0$. Then,
$$\det (B) = \det (A\cdot -v^T A^{-1} u)=(-v^T A^{-1} u)\cdot \det(A) $$
Because $A$ is nonsingular, we know that $\det(A)\neq 0$ and moreover, that $A^{-1}u\neq0$ for $u\neq 0$.
My Question: How do I show that $v^T A^{-1} u\neq 0$?
You did everything right (up to the computation of the determinant, which is not written down correctly). Doing Laplace expansion on the last row, $$ \det B=(-v^TA^{-1}u)\det A $$ so this is nonzero if and only if $v^TA^{-1}u\ne0$.
You can't prove that $v^TA^{-1}u\ne0$, because it's false in general (think to $u=0$, for instance), but the question was asking precisely to show an “if and only if” statement.