Here is the second installment that answers the converse in the affirmative. The result is not easy to establish. One method of proof uses the spectral calculus for self-adjoint operators, but this is like cracking a nut with a sledgehammer. I provide a softer approach below, which exploits the geometric properties of Hilbert spaces.
Lemma 1 Every bounded sequence in a Hilbert space contains a weakly convergent subsequence.
Proof This follows from the reflexivity of Hilbert spaces. Q.E.D.
Lemma 2 Every weakly convergent sequence in a Hilbert space is bounded.
Proof This follows from the Uniform Boundedness Principle. Q.E.D.
Definition 1 Let $ \mathcal{H} $ be a Hilbert space, and let $ C $ be a fixed collection of sequences in $ \mathcal{H} $. Given a sequence $ (\mathbf{x}_{n})_{n \in \mathbb{N}} $ in $ \mathcal{H} $, we say that $ (\mathbf{x}_{n})_{n \in \mathbb{N}} $ can be approximated by $ C $ if for every sequence $ (\epsilon_{n})_{n \in \mathbb{N}} $ of positive real numbers, there exists a $ (\mathbf{c}_{n})_{n \in \mathbb{N}} \in C $ such that $ \| \mathbf{c}_{n} - \mathbf{x}_{n} \|_{\mathcal{H}} < \epsilon_{n} $ for all $ n \in \mathbb{N} $.
Definition 2 Let $ \mathcal{H} $ be a Hilbert space. We denote by $ \mathbf{BOS}(\mathcal{H}) $ the set of all bounded orthogonal sequences in $ \mathcal{H} $.
Lemma 3 Let $ \mathcal{H} $ be a Hilbert space, and let $ (\mathbf{x}_{n})_{n \in \mathbb{N}} $ be a weak null-sequence in $ \mathcal{H} $. Then $ (\mathbf{x}_{n})_{n \in \mathbb{N}} $ contains a subsequence that can be approximated by $ \mathbf{BOS}(\mathcal{H}) $.
Proof Let $ (\mathbf{x}_{n})_{n \in \mathbb{N}} $ be a weak null-sequence in $ \mathcal{H} $. Fix a sequence $ (\epsilon_{n})_{n \in \mathbb{N}} $ of positive real numbers. We inductively define a new sequence $ (\mathbf{v}_{n})_{n \in \mathbb{N}} $ in $ \mathcal{H} $ and an increasing sequence $ (\alpha_{n})_{n \in \mathbb{N}} $ of positive integers as follows:
Set $ \alpha_{1} := 1 $ and $ \mathbf{v}_{1} := \mathbf{x}_{1} $.
For each $ n \in \mathbb{N} $, suppose that $ \alpha_{1},\ldots,\alpha_{n} $ and $ \mathbf{v}_{1},\ldots,\mathbf{v}_{n} $ have been defined. As $ (\mathbf{x}_{n})_{n \in \mathbb{N}} $ converges weakly to $ 0_{\mathcal{H}} $, we can choose a smallest positive integer $ k > \alpha_{n} $ such that \begin{equation} \left\| \sum_{i=1}^{n} \lambda_{i} \mathbf{v}_{i} \right\|_{\mathcal{H}} < \epsilon_{n}, \end{equation} where \begin{equation} \lambda_{i} = \left\{ \begin{array}{ll} \dfrac{\langle \mathbf{x}_{k},\mathbf{v}_{i} \rangle}{\| \mathbf{v}_{i} \|_{\mathcal{H}}^{2}} &\text{, if $ \| \mathbf{v}_{i} \|_{\mathcal{H}} > 0 $}; \\ 0 &\text{, if $ \| \mathbf{v}_{i} \|_{\mathcal{H}} = 0 $}. \end{array} \right. \end{equation} Then set \begin{equation} \alpha_{n+1} := k \quad \text{and} \quad \mathbf{v}_{n+1} := \mathbf{x}_{k} - \sum_{i=1}^{n} \lambda_{i} \mathbf{v}_{i}. \end{equation}
Notice that $ (\mathbf{v}_{n})_{n \in \mathbb{N}} $ is the result of applying the Gram-Schmidt orthogonalization procedure to $ (\mathbf{x}_{\alpha_{n}})_{n \in \mathbb{N}} $, which is a subsequence of $ (\mathbf{x}_{n})_{n \in \mathbb{N}} $. Therefore, $ (\mathbf{v}_{n})_{n \in \mathbb{N}} $ is an orthogonal sequence. By Lemma 2, $ (\mathbf{x}_{n})_{n \in \mathbb{N}} $ is bounded, so $ (\mathbf{v}_{n})_{n \in \mathbb{N}} \in \mathbf{BOS}(\mathcal{H}) $. Finally, $ \| \mathbf{v}_{n} - \mathbf{x}_{\alpha_{n}} \|_{\mathcal{H}} < \epsilon_{n} $ for all $ n \in \mathbb{N} $. Q.E.D.
Theorem Let $ \mathcal{H} $ and $ \mathcal{K} $ be Hilbert spaces. Let $ T: \mathcal{H} \rightarrow \mathcal{K} $ be a bounded linear operator that maps every orthonormal sequence (hence every bounded orthogonal sequence) in $ \mathcal{H} $ to a strong null-sequence in $ \mathcal{K} $. Then $ T $ is a compact operator.
Proof Let $ (\mathbf{x}_{n})_{n \in \mathbb{N}} $ be a bounded sequence in $ \mathcal{H} $. By Lemma 1, there exists a weakly convergent subsequence $ (\mathbf{x}_{n_{k}})_{k \in \mathbb{N}} $ of $ (\mathbf{x}_{n})_{n \in \mathbb{N}} $. Let $ \mathbf{x} $ be the weak limit of this subsequence. Clearly, $ (\mathbf{x}_{n_{k}} - \mathbf{x})_{k \in \mathbb{N}} $ is then a weak null-sequence in $ \mathcal{H} $. By Lemma 3, there exists a subsequence $ (\mathbf{x}_{n_{k_{l}}} - \mathbf{x})_{l \in \mathbb{N}} $ of $ (\mathbf{x}_{n_{k}} - \mathbf{x})_{k \in \mathbb{N}} $ and a sequence $ (\mathbf{v}_{l})_{l \in \mathbb{N}} \in \mathbf{BOS}(\mathcal{H}) $ such that \begin{equation} \forall l \in \mathbb{N}: \quad \| \mathbf{v}_{l} - (\mathbf{x}_{n_{k_{l}}} - \mathbf{x}) \|_{\mathcal{H}} < \frac{1}{l}. \end{equation} Observe that $ T $ must map $ (\mathbf{v}_{l})_{l \in \mathbb{N}} $ to a strong null-sequence in $ \mathcal{K} $. Hence, by the approximation property, we have $ \displaystyle \lim_{l \rightarrow \infty} T(\mathbf{x}_{n_{k_{l}}} - \mathbf{x}) = 0_{\mathcal{K}} $. In other words, $ (T(\mathbf{x}_{n}))_{n \in \mathbb{N}} $ contains $ (T(\mathbf{x}_{n_{k_{l}}}))_{l \in \mathbb{N}} $ as a strongly convergent subsequence. Therefore, $ T $ is a compact operator. Q.E.D.