= Solution
The exact condition for this <diagonal operator on sequence space> is
$$
\boxed{T\text{ is compact on }\ell^2\quad\Longleftrightarrow\quad\lambda_n\longrightarrow0.}
$$
First, $T$ is bounded precisely when $\sup_n|\lambda_n|<\infty$, with <operator norm> equal to that supremum: the upper bound follows by summing $|\lambda_nx_n|^2$, and the coordinate unit vectors $e_n$ give the lower bound. If the supremum is infinite, choose distinct indices with $|\lambda_{n_j}|\ge j$ and set $x_{n_j}=1/j$, other coordinates zero. Then $x\in\ell^2$ but $Tx\notin\ell^2$, so the displayed prescription would not even define an everywhere-defined map into $\ell^2$.
If $T$ is compact but $\lambda_n\not\to0$, infinitely many indices satisfy $|\lambda_n|\ge\epsilon>0$. Along those indices,
$$
\|Te_n-Te_m\|^2=|\lambda_n|^2+|\lambda_m|^2\ge2\epsilon^2\quad(n\ne m),
$$
contradicting <compactness>. Conversely, if $\lambda_n\to0$, truncate to $T_Nx=(\lambda_1x_1,\ldots,\lambda_Nx_N,0,\ldots)$. This is a finite-rank <compact operator>, and $\|T-T_N\|=\sup_{n>N}|\lambda_n|\to0$. To see <compactness> directly, approximate the image of the <unit ball> by $T_N$ to within $\epsilon/2$, and cover its bounded finite-dimensional image by finitely many $\epsilon/2$-balls. Hence $T$'s unit-ball image is <totally bounded>; its closure is compact because $\ell^2$ is complete.
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