Let $\mathcal{H}$ be a separable Hilbert space. Let $(\mathcal{S}_\alpha)_{\alpha\in\Lambda}$ be a decreasing net of closed subspaces of $\mathcal{H}$, i.e. such that for each $\alpha,\beta\in\Lambda$, there exist $\gamma\ge\alpha,\beta$ such that $\mathcal{S}_\gamma\subseteq \mathcal{S}_\alpha\cap \mathcal{S}_\beta$.
Using separability of $\mathcal{H}$, can we conclude that $(\mathcal{S}_\alpha)$ admits a subnet of countable cardinality?
No. For example, consider a free ultrafilter $\mathcal{U}$ on $\mathbb{N}$. For each subset $A$ of $\mathbb{N}$, it corresponds to a subset of the standard basis of $\ell^2$ and therefore corresponds to a closed subspace of $\ell^2$. Thus, $\mathcal{U}$ induces a decreasing net of closed subspaces of $\ell^2$. There can be no countable subnet because if such a countable subnet exists, then $\mathcal{U}$, as a point in $\beta\mathbb{N}$, would admit a countable local base, which is impossible. See https://mathoverflow.net/questions/464384/points-in-the-stone-cech-compactification-are-intersection-of-open-sets.