Prove that there exists some $k $ with $1\leq k\leq n$ such that $C_{kk}\neq 0$ where $C_{kk}$ is the cofactor of $a_{kk}$ where $A= (a_{ij})$ is a symmetric matrix of rank $n-1.$
First approach: Suppose not. Let $\text{rank}(A) = n-1.$ Then all the diagonal entries of $B =\text{adj}A$ are $0$. Since $AB= 0$ (as $A\times \text{adj}A= \det A \times I_n$ and $\det A= 0$ as $A$ is not full rank), $\text{rank} AB=0$ so by Sylvester's rank inequality $(\text{rank}A + \text{rank}B -n \leq \text{rank}AB)$ it follows that $\text{rank} B \leq 1.$ If $\text{rank} B=0$ then $B=0$ contradicting the fact that since $\text{rank}A = n-1$ there is some $(n-1)\times (n-1)$ submatrix with non-zero determinant. Therefore $\text{rank} B=1.$ Since $A$ is symmetric, $B$ is symmetric so $B$ must be of the form
$B= \left(\begin{matrix}0&&x_1 &&\cdots &&x_{n-1}\\x_1&&0&&\cdots &&x_{n-2}\\\vdots&& &&\ddots &&\\x_{n-1}&&x_{n-2}&& \cdots&&0 \end{matrix}\right)$
Since $\text{rank}B =1$ we can assume WLOG that rows $2$ to $n$ are scalar multiples of row $1$ but this will imply that $x_i=0$ for $i=1, \ldots n-1$ implying that $B=0$ which we already showed leads to a contradiction.
My first question: is this approach right?
My second question is: can the second approach below be salvaged somehow?
Second approach: Since $A$ is real and symmetric it is orthogonally diagonalisable and since its rank is $n-1$ it has $0$ as an eigenvalue with algebraic (and hence geometric) multiplicity $1$. If $A = PDP^{T}$ where $D =\text{diag}(d_{11}, \ldots, d_{nn})$ then there is some $k$ such that $d_{kk}=0.$ Can we say something about the rank of $A'$ where $A'$ is obtained from $A$ by deleting the $k^{th}$ row and column? (Again we know it is orthogonally diagonalisable but can we say for sure that $A'$ does not have $0$ as an eigenvalue? This might be a little farfetched, but can the interlacing inequalities of eigenvalues of principal submatrices of Hermitian matrices be applied here?)
Your first proof looks fine to me. I am not sure about the second one, but you may follow a third approach: the sum of all principal $(n-1)\times(n-1)$ minors of $A$ is $\sum_i\prod_{j\ne i}\lambda_j$.