Question

Let H be an infinite dimensional and separable Hilbert space. Let C be a closed and connected subset of H containing more than one point. Can C ever be the countable union of closed and totally disconnected subsets of H?

Answer 1

Let $E_c$ be the complete Erdös space (Erdös, Annals of Math vol 41 1940), defined as the subspace of $\ell^2(\mathbb{N})$ where all coordinates are irrationals. It is polish (separable and completely metrizable) and totally disconnected, but admits a "connectification" namely a (still polish) topology on $E_c\cup\{p\}$ that makes it connected (and of course induces the one on $E_c$). The crucial point is the fact that any nonempty closed and open subset of $E_c$ is unbounded. Then, as in Bill Johnson's answer $E_c$ is the union of the closed and totally disconnected subspaces $\overline{B}(0,n)\cup\{p\}$, $n\geq 1$. It remains to remark that, like any polish space, $E_c\cup\{p\}$ embeds as a closed subset of $H$ (as remarked in Gerald Edgar's answer).

Answer 2

$C$ is simply a connected complete separable metric space with more than one point. Now the union of countably many closed sets of topological dimension zero must again have dimension zero. But I suppose "totally disconnected" is not quite the same as "zero dimensional" so this is not yet a complete answer.

Answer 3

Let $C$ be an explosion point space such as the Knaster–Kuratowski fan (http://en.wikipedia.org/wiki/Knaster–Kuratowski_fan) and $p$ the explosion point in $C$.
Set $A_n = \{p\}\cup (C\sim B(p; 1/n))$.