Let $X$ be a continuum $=$ a connected compact metric space.
Define $x\sim y$ if $x$ and $y$ are contained in a nowhere dense subcontinuum of $X$. It is easy to see that $\sim$ is an equivalence relation.
Examples:
$|[0,1]/\sim|=|\mathbb R|$
$|[0,1]^2/\sim|=1$
Question: Is there a continuum $X$ with $1<|X/\sim|<|\mathbb R|$?
This is an incomplete answer (this question is non-trivial, it is actually quite difficult IMO) but I thought I'd put it here with the hope that someone could complete the proof (I've spent a good amount of time on it already and unfortunately have to move on).
$\textbf{Proposition:}$ Suppose $X$ is a path-connected continuum. Then under the equivalence relation previously described, there are either infinitely many equivalence classes or there is only one.
$\textbf{Proof:}$ There are two cases here:
1.) If for each $x \neq y \in X$, there is a path $f: [0,1] \rightarrow X$ between $x$ and $y$ that is nowhere dense in $X$, then the closure $\overline{f([0,1])} \subseteq X$ is a nowhere dense, connected (closure of continuous image of connected set), compact (closed subset of compact space) set and hence a nowhere dense subcontinuum. Thus, in this case, we have $x \sim y$ for each pair $x \neq y$ and as a result only one equivalence class.
2.) Now suppose we have $x \neq y \in X$ such that every path $f:[0,1] \rightarrow X$ between them is dense in some nonempty open set. As a metric space, $X$ is Hausdorff and so path connectedness implies arc-connectedness, so there is a path between $x$ and $y$, $f:[0,1] \rightarrow f([0,1]) \subseteq X$, that is also a homeomorphism. By hypothesis, there is a non-empty open ball $U \subseteq X$ such that $f([0,1]) \cap U =V$ is dense in $U$. We now argue that $U \cong (a,b) \subseteq \mathbb{R}$ for some $a<b$.
Consider $U$ as a subspace of $X$. Note that $f([0,1]) \subseteq X$ is closed and thus by definition of the subspace topology, $V$ is closed. We then define the map $F: U \rightarrow \mathbb{R}$, $F: x \mapsto d(x, V)$, where $d$ is the metric defined on $X$. Since $V$ is closed, $F$ is continuous. But note that for each $x \in V$, $F(x)=0$, so $F$ agrees with the zero function on a dense subset of a Hausdorff space and so must be $0$ everywhere. Then $F(U)= \left\{0 \right\}$, i.e. for every $x \in U$, $d(x, V)=0$. But then $U=V$. Then since $U$ is an open ball, it is connected and its image under the homeomorphism $f^{-1}: f([0,1]) \rightarrow [0,1]$ must be connected and open, and so is an interval of the form $(a,b)$. Then $U \cong (a,b)$ as claimed. (Obviously, $U \neq \emptyset$ implies $a<b$)
$\textit{Lemma:}$ Suppose $S \subseteq X$ is a connected set such that there are two distinct points $c,d \in S \cap U$. Then $S \cap U$ is connected.
(presented without proof, I haven't been able to prove this lemma).
If we suppose we have $c,d \in U$ such that $c\sim d$, then there is a connected, nowhere dense set $S$ with $c,d \in S$. Then by the lemma, $S \cap U$ is connected and since it contains at least two points, its image under the homeomorphism $f^{-1}$ is an interval. But then $S$ is dense in some open set in $U$, contradicting the "nowhere dense" hypothesis. We then conclude that given two points $c,d \in U$, they cannot be elements of the same equivalence class. It then follows that there are infinitely many classes, since $U$ has infinitely many elements. This completes the proof. $\Box$
So it comes down to that lemma (assuming everything else is OK). It seems intuitively true but I haven't actually been able to prove it. Also, I can't see any immediate way to extend this to the case where the space is just connected.