How thinly connected can a closed subset of Hilbert space be? Let H be a separable (and infinite-dimensional) Hilbert space. Is it known whether there exists an infinite
subset C of H with the following properties.? (1) C is connected and closed in H. (2) No infinite proper 
subset of C is both connected and closed in H.
Perhaps the answer depends upon whether or not the Axiom of Choice is assumed to hold. In any case, no 
finite-dimensional Euclidean space-because it is locally compact-can contain a subset with these properties.
 A: I solved this problem in  2018 in this paper on arXiv (published version here).
There is a Polish space (a closed subspace of Hilbert space) that is connected, but has no proper closed connected subsets other than singletons.  The term used in the paper and elsewhere is widely-connected.

Surely continuum theorists have thought about this question, haven't they?

Indeed this problem appeared in the Houston Problem Book and various other lists of problems by continuum theorists. But it was first asked by Paul Erdős.
EDIT: In response to @SamHopkins comment, I will outline the construction here. Begin with the bucket-handle continuum $K$:

There is a compact set $D\subset  K$ such that $D$ intersects each arc-component of $K$ and $K\setminus D$ is connected (shown here). I proved that a countably infinite  collection $D_1,D_2\ldots$ of copies of $D$ can be removed from $K$ so that the remainder $X=K\setminus \bigcup_{i=1}^\infty D_n$ is connected, and each arc in $K$ intersects some $D_n$. The space $X$ is Polish, and since each proper subcontinuum of $K$ is an arc it follows that $X$ is widely-connected.
A: Some remarks, not an answer.

*

*As pointed out by BS, the problem is: does there exist a connected Polish space containing no nondegenerate proper closed connected subspace? In the language of continuum theory this sounds suspiciously simple: does there exist a connected Polish space whose all composants are singletons? Or (still equivalently): does there exist a connected Polish space that is irreducible between every pair of distinct points? Surely continuum theorists have thought about this question, haven't they?


*As hinted by Garabed, such a space $X$ cannot be locally compact. Indeed, in locally compact spaces, components coincide with quasi-components. Let $F$ be a closed ball of some radius about some $x\in X$, so that $F\ne X$, and let $C$ be the component of $x$ in $F$.
If $C\ne\{x\}$, then $C$ is a closed connected nondegenerate subset of $X$. If $C=\{x\}$, then $\{x\}$ is also a quasi-component, so $x$ is contained in arbitrarily small clopen sets in $F$. Then they are also clopen in $X$, so $X$ cannot be connected.


*There exists a connected Polish space $X$ containing no nondegenerate compact connected subspace. For instance, the graph of $f(x)=\sum_{n=1}^\infty 2^{-n}\sin(\frac1x-r_n)$, where $r_n$ is the $n$th rational number (in some order) and $\sin(\infty)=0$. See Kuratowski's "Topology" (volume II, section 47.IX in the 1968 edition).
In fact, such an $X$ can even be locally connected (of course, it cannot be locally path-connected at any point).


*The graph of a discontinuous function $f:\Bbb R\to\Bbb R$ satisfying $f(x+y)=f(x)+f(y)$
can be connected, and in that case it contains no nondegenerate bounded connected subset. (Reference)
A: This is not really an answer, just remarks on the problem, but it would be too long for a comment. Apologies.
As remarked by François Dorais in a comment, you can phrase the question entirely in terms of polish spaces (i.e. completely metrizable and separable topological spaces). And since a connected metric space with at least two points is infinite, you are asking if there exists an infinite connected polish space such that any proper closed subspace is totally disconnected (equivalently, any proper closed connected subset is a point, or empty).
A first (failed) attempt to an example would be the complete Erdös space $E_c$ (Erdös, Annals of Math vol 41 1940), defined as the subspace of $\ell^2(\mathbb{N})$ where all coordinates are irrationals. It is polish and totally disconnected, but admits a connectification namely a (still polish) topology on $E_c\cup\{\infty\}$ that makes it connected (and of course induces the one on $E_c$). So it is rather "thinly" connected, but maybe not in your sense.
Another amazing property of $E_c$ is that it is homeomorphic to the subspace of $\ell^2(\mathbb{N})$ where all coordinates belong to $\{0\} \cup \{1/n\}_{n\geq 1}$. This sounds rather improbable. 
All this (and much more) is explained in papers on JJ Dijkstra publications page
e.g. 27, 30 and 32.
