Proving there is an eigenvalue $\lambda$ for which $|\lambda - b_{jj}| < \epsilon \sqrt{n}$

The eigenvalues of a matrix and its similarity transform are the same, so the eigenvalues of $A$ and $B$ are the same.

Next, for every $j=1,2,\ldots,n$, define a symmetric matrix $E^{(j)}=(B-b_{jj}I)e_je_j^T+e_je_j^T(B-b_{jj}I)$, where $e_j$ is $j^\text{th}$ standard basis vector. We have \begin{equation} (B-E^{(j)})e_j= Be_j - (B-b_{jj}I)e_j+e_je_j^T(B-b_{jj}I) e_j= b_{jj}e_j, \end{equation} since $e_je_j^TBe_j = e_j(e_j^TBe_j )= b_{jj}e_j.$ Thus, $b_{jj}$ is an eigenvalue of $B-E^{(j)}$, and hence we invoke Bauer-Fike theorem to show that there exists an eigenvalue $\lambda$ of $B$ such that \begin{equation} \vert b_{jj}-\lambda\vert\leq \Vert E^{(j)}\Vert = \Vert(B-b_{jj}I)e_j\Vert = \sqrt{\sum_{i=1,i\neq j}^nb_{ij}^2}\leq \sqrt{n-1}\epsilon<\sqrt{n}\epsilon. \end{equation}