= Solution
Normalize the mutually orthogonal eigenvectors so that $(x_j)$ is an <orthonormal sequence>. If $\lambda_j$ did not tend to zero, some subsequence would satisfy $|\lambda_j|\geq\varepsilon>0$. For distinct indices in this subsequence, orthogonality gives
$$
\lVert Tx_j-Tx_k\rVert^2
=\lVert\lambda_jx_j-\lambda_kx_k\rVert^2
=|\lambda_j|^2+|\lambda_k|^2
\geq2\varepsilon^2.
$$
Thus $(Tx_j)$ would have no Cauchy subsequence, contradicting compactness of $T$. Therefore the <orthogonal eigenvectors of a compact operator> satisfy
$$
\boxed{\lambda_j\longrightarrow0.}
$$
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