Let $H$ be Hilbert space and $K\in L(H,H)$ be a compact operator. Show that $T=K^*K$ is compact and symmetric. Then use the Hilbert-Schmidt Theorem to show that there is an orthonormal sequence $\{\varphi_k\}$ of $H$ such that $T(\varphi_k)=\lambda_k \varphi_k$ for all $k$ and $T(h)=0$ if $h$ is orthogonal to $\{\varphi_k\}$. Conclude that if $h$ is orthogonal to $\{\varphi_k\}$, then $K(h)=0$.
Define $H_0$ to be the closed linear span of $\{K^m(\varphi_k),m\ge 1,k\ge 1\}$, Show that $H_0$ is closed and separable, $K(H_0)\subset H_0$ and $K=0$ on $H_0^\perp$.
I find some difficulty in proving the last result that $K=0$ on $H_0^\perp$ despite the remaining part of this problem being solvable for me.
As Chrystomath noted, for any $h \in H_0^\perp$: $\langle h, K(\phi_k) \rangle =\langle K^*(h),\phi_k \rangle = 0$ for all $k$, so $K^*(h)$ is orthogonal to $\{\phi_k\}$. Thus $K(K^*(h)) = 0$. Then $||K^*(h)||^2 = \langle K^*(h), K^*(h) \rangle = \langle KK^*(h), h \rangle = 0$, so we have $K^*(h) =0$.