prove that $|\lambda(H) - \lambda(B)| \leq \sqrt{||(C^HC)||_2}$ Let A, B be Hermitian square matrices and 
$$H = \left[\begin{array}{rr}A & C \\ C^H & B\end{array}\right]$$
Show every eigenvalue $\lambda(B)$ of B, there is an eigenvalue $\lambda(H)$ of H such that $$|\lambda(H) - \lambda(B)| \leq \sqrt{||C^HC||_2}$$
$||\cdot||_2$ means 2-norm
Can someone explain to me how to do this problem, because I can only think of determinant of $det(H - \lambda I)$, but it does not make sense in this problem. 

This is just a simple consequence of the Bauer-Fike theorem. Let $M=XDX^{-1}$ be a diagonalizable matrix. The theorem says that for any eigenvalue $\lambda$ of $M$ there is an eigenvalue $\mu$ of $M+N$ such that $|\lambda-\mu|\leq\kappa_2(X)\|N\|_2$, where $\kappa_2(X)$ is the spectral condition number of the eigenvector matrix $X$. If $M$ is normal (which includes Hermitian matrices), $X$ can be chosen orthogonal so $\kappa_2(X)=1$.
Applying this to the Hermitian matrix $H$, for any eigenvalue $\lambda$ of $A$ or $B$, that is, of
$$
\pmatrix{A&0\\0&B}=:M,
$$
there is an eigenvalue $\mu$ of 
$$
H=\pmatrix{A&0\\0&B}+\pmatrix{0&C\\C^*&0}=:M+N
$$
such that
$$
|\lambda-\mu|\leq\left\|\pmatrix{0&C\\C^*&0}\right\|_2.
$$
The norm of $N$ is equal to the norm of $C$ since
$$
N^*N=\pmatrix{0&C\\C^*&0}\pmatrix{0&C\\C^*&0}=\pmatrix{CC^*&0\\0&C^*C},
$$
so $\|N\|_2=\rho(N^*N)^{1/2}=\max\{\rho(CC^*)^{1/2},\rho(C^*C)^{1/2}\}=\|C\|_2$.
