Implicit function theorem for eigenvalues: why simple eigenvalue matters?

Notice that if $\lambda_0$ is not simple, then we can take another eigenvector corresponding to $\lambda_0$ and make it orthogonal to $u_0$. Call this vector $v$. Then we have $$ \begin{pmatrix}\lambda_0I-X_0 & u_0\\ 2u_0^T&0\end{pmatrix} \begin{pmatrix}v\\0\end{pmatrix} = \begin{pmatrix}\lambda_0v - X_0v + 0 \\ 2u_0^Tv + 0\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}, $$ So the matrix is no longer invertible. I'm not going to proof the iff statement, but this is why it matters.

To verify the claim about the value of the determinant, complete the eigenbasis of $X_0$ to obtain $\{u_0, v_1,\dots,v_{n-1}\}$ corresponding to eigenvalues $\lambda_0,\lambda_1,\dots\lambda_n$. Since $X_0$ is symmetric, $u_0\perp v_i$. We then have $$ \begin{pmatrix}\lambda_0I-X_0 & u_0\\ 2u_0^T&0\end{pmatrix} \begin{pmatrix}v_i\\0\end{pmatrix} = \begin{pmatrix}\lambda_0v_i - X_0v_i + 0 \\ 2u_0^Tv_i + 0\end{pmatrix} = (\lambda_0 - \lambda_i)\begin{pmatrix}v_i\\0\end{pmatrix}, $$ so we have that $\lambda_0-\lambda_1,\dots\lambda_0-\lambda_{n-1}$ are eigenvalues of the block matrix, and these are also the eigenvalues of $\lambda_0I-X_0$. Note that this is the matrix used in the claim in the linked proof, not $X_0$. This has given us $n-1$ of the eigenvalues of the block matrix.

To find the last two, notice that if we normalize $\|u_0\|=1$ , then the vectors $(u_0, \sqrt{2})^T$ and $(u_0,-\sqrt{2})^T$ are eigenvectors of the block matrix with eigenvalues $\sqrt{2}$ and $-\sqrt{2}$, respectively. Since the determinant is the product of the eigenvalues, we have that $\det A = \left(\prod_{i=1}^{n-1} \lambda_0-\lambda_i\right)(\sqrt{2})(-\sqrt{2})$, which is $-2$ times the product of the nonzero eigenvalues of $\lambda_0 I - X_0$, as desired.