For notational simplicity I will take the index set to be $\mathbb N$ instead of $\mathbb Z$ (it doesn't change the substance of the proof).
Fix $f\in\mathcal H$. Consider the rank-one operators $E_k=\langle\cdot,\eta_k\rangle\,\eta_k$. Then $\tag{1} \left\|\sum_{k=1}^nE_kf\right\|^2=\left\langle\sum_{k=1}^nE_kf,\sum_{j=1}^nE_jf\right\rangle=\sum_{j,k=1}^n\langle E_jE_kf,f\rangle=\sum_{j,k=1}^n\langle f,\eta_j\rangle\,\langle\eta_k,\eta_j\rangle\,\langle\eta_k,f\rangle. $ Let $g_n\in\ell^2(\mathbb N)$ be given by $ g_n(k)=\begin{cases}\langle f,\eta_k\rangle,&\text{ if }k\leq n \\ 0,&\text{ otherwise} \end{cases} $ Then, from (1), $ \left\|\sum_{k=1}^nE_kf\right\|^2=\langle Ag_n,g_n\rangle\leq \|A\|\,\|g_n\|^2. $ So $ \sum_{k=1}^n|\langle f,\eta_k\rangle|^2=\sum_{k=1}^n\langle f,\eta_k\rangle\,\langle \eta_k,f\rangle=\sum_{k=1}^n\langle E_kf,f\rangle=\langle\sum_{k=1}^n E_kf,f\rangle\leq\left\|\sum_{k=1}^nE_kf\right\|\,\|f\|\\ \leq \|A\|^{1/2}\,\|g_n\|\,\|f\| =\|A\|^{1/2}\,\|f\|\,\left(\sum_{k=1}^n|\langle f,\eta_k\rangle|^2\right)^{1/2}. $ After dividing by $\left(\sum_{k=1}^n|\langle f,\eta_k\rangle|^2\right)^{1/2}$ and squaring, we get $ \sum_{k=1}^n|\langle f,\eta_k\rangle|^2\leq\|A\|\,\|f\|^2. $ As this holds for any $n$, $ \sum_{k=1}^\infty|\langle f,\eta_k\rangle|^2\leq\|A\|\,\|f\|^2. $