= Solution
Suppose the <compact operator> $K$ acts on $H$, and $h_n\rightharpoonup h$. The <Uniform boundedness principle> makes $(h_n)$ bounded, so its images form a relatively compact set. Bounded linearity gives $Kh_n\rightharpoonup Kh$: for every $g$, $\langle Kh_n,g\rangle=\langle h_n,K^*g\rangle\to\langle Kh,g\rangle$. Every norm-convergent subsequence of $(Kh_n)$ must therefore converge to $Kh$. If the whole sequence did not converge to $Kh$ in norm, a subsequence staying a fixed positive distance from $Kh$ would have a norm-convergent further subsequence, a contradiction. This proves that <compact operators send weak convergence to norm convergence>.
Conversely, assume the stated complete continuity property. Its strong limit necessarily equals $Kh$, because bounded linearity still supplies the weak limit. Every bounded sequence in a <Hilbert space> has a weakly convergent subsequence by <weak sequential compactness in a Hilbert space>. Applying the assumption to that subsequence gives a norm-convergent subsequence of the images. Thus the image of the <unit ball> is relatively sequentially compact and hence relatively compact. \b[Complete continuity and compactness are equivalent on a Hilbert space.]
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