Suppose $f_k$ is an orthonormal basis of a separable Hilbert space. $T$ is bounded. $\| (T-\lambda_k)f_k \| \rightarrow 0$, $\| (T-\lambda_k)^*f_k \| \rightarrow 0$. $R$ is the diagonal operator with eigenvalues and eigenvectors $\lambda_k$ and $f_k$ respectively. To prove $T-R$ is compact.
My attempt:
We are given $\| (T-R)f_k \| \rightarrow 0$, $\| (T-R)^*f_k \| \rightarrow 0$.
$R_k(f_i):=\begin{cases}R(f_i)& i\leq k\\ T(f_i)& \text{otherwise}\end{cases}$
So, $T-R_k\rightarrow T-R$. I am not sure if $T-R_k$ is finite rank. Also I have not used $\| (T-\lambda_k)f_k \| \rightarrow 0$, $\| (T-\lambda_k)^*f_k \| \rightarrow 0$.Another criterion is
$T \in \mathcal{L}(X,Y)$ satisfies that $\|Tx_n − Tx\| \rightarrow 0$, as $n \rightarrow \infty$, whenever $(x_n)_{n\geq 1}$ is a sequence in $X$ converging weakly to $x \in X$, then $T \in \mathcal{K}(X, Y )$.
Any help is appreciated.
Edit: The space is separable.