= Solution
Compactness gives a stronger statement than the one requested. Since $T(B_H)$ has compact closure, finitely many open balls of radius $\delta$ cover it. Choose their centres from $T(B_H)$ and let $E$ be their finite-dimensional linear span. Then
$$
\sup_{\|x\|\leq1}\operatorname{dist}(Tx,E)<\delta.
$$
For the <orthogonal projection> $P_E$, best approximation in a Hilbert space gives
$$
\|(I-P_E)Tx\|<\delta
\qquad(\|x\|\leq1).
$$
In particular this holds for every vector $e_n$ in the given <orthonormal basis>.
The operator $P_ET$ has image in $E$ and hence has <finite-rank operator>[finite rank], while
$$
\|T-P_ET\|
=\sup_{\|x\|\leq1}\|(I-P_E)Tx\|<\delta.
$$
\b[Thus every compact operator is an <operator norm> limit of finite-rank operators. This is the <finite-rank approximation theorem for compact operators on a Hilbert space>.]
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