I hope that my example below is as elegant as the continuous long line provided by Goldstern above, while my example is less expected. Also, while long line is simpler **in itself**, the **proof** is simpler in my case. Finally, perhaps logicians will find some advantages (I'll do a little of it--I am not confident to do it well).

Let $\ A\ $ be an arbitrary set. The following ordered triple $\ (\mathbf S_A\ \mathbf 0\ \mathbf 1)\ $, where $\ \mathbf S_A:=(S_A\ T_A)\ $ is a topological space--call it a skeleton, and $\ \mathbf {0\ 1}\in S_A,\, $ is to be defined below, while first (ahead of time) let's formulate

**THEOREM** For every connected subset $\ X\subseteq S_A,\ $ such that $\ \mathbf {0\ 1}\in X,\ $ the inequality of cardinalities $\ |X|\ge|A|\ $ holds.

This instantly gives a simple negative answer to *the question of this thread* posed by Dominic.

**DEFINITION**

- $\ S_A\ :=\ \{(x_a)_{a\in A}\in[0;1]^A\ :\ \forall_{a\ b\in A} [x_a\ x_b\in(0;1)\ \Rightarrow\ a=b]\ \}$
- $\ \mathbf 0\ :=\ (0)_{a\in A}\ $ and $\ \mathbf 1\ :=\ (1)_{a\in A}$
- $\ T_A\ $ is the topology in $\ S_A\ $ induced by the Tikhonov toplogy in cube $\ [0;1]^A$

**PROOF (of the theorem)** The connected component of $\ \mathbf 0\ $ in $\ S_A\ $ is dense in $\ S_A,\ $ which means that its closure, i.e. space $\ S_A\ $ itself, is connected too. Next, let:

$$H^a\ :=\ \{x\in[0;1]^A\ :\ x_a=\frac 12$$

Let $\ X\subseteq S_A\ $ be a connected subset such that $\ \mathbf {0\ 1}\in X.\ $ Then $\ p_a(X)=[0;1],\ $ hence $\ H\,^a\cap X\ne \emptyset.\ $ Thus

$$ |X|\ \ge\ \left|\left\{H^a\ :\ a\in A\ \right\}\right|\ =\ |A|$$

Indeed, sets $\ H^a\ $ are disjoint (hence different). **END of Proof**

G E N E R A L I Z A T I O N

We may replace the topological interval $\ [0;1],\ $ and its three points $\ 0\ \frac 12\ 1,\ $ by an arbitrary connected space $\ S\ $, and its three points $\ a\ h\ b,\ $ such that $\ h\ $ separates $\ a\ b\ $ (meaning that there are open sets $\ G\ $ and $\ H:=(S\setminus\{h\})\setminus G\ $ of $\ S\ $ such that $\ a\in G\ $ and $\ b\in H$. Etc. The theorem still holds.

Logical considerations

I am not using ordinal numbers. My construction is free of any complications, especially when $\ S\ $ of the generalization is a proper 3-point space $\ \{a\ h\ b\}.\ $ Thus I am worried only about axioms like the axiom of choice or continuum hypothesis, and similar, about their relation to the cartesian product, and to the ordinary $\ [0;1]\ $ of my main example.

EXTRA. Compactness. Another connectedness proof.

Space $\ \mathbf S_A\ $ is compact, it is a closed subset of the Tikhonov cube $\ [0;1]^A:\ $ **indeed**, let $\ x\in [0;1]^A\setminus S_A.\ $ Then there exist two **different** indices $\ a\ b\ \in\ A\ $ such that $\ (x_a\ x_b)\in (0;1)^{\{a\ b\}}.\ $ Thus the inverse image of this open square under the canonical projection $\ p_{a\ b} : [0;1]^A\rightarrow(0;1)^{\{a\ b\}}\ $ is disjoint from $\ S_A\ $ (one could say that $\ [0;1]^A\setminus S_A\ $ is open because it is a union of the inverse images of the open squares). Thus indeed $\ S_A\ $ is compact.

Now $\ \mathbf S_A\ $ is connected because it is an inverse limit of spaces $\ \mathbf S_B\ $ for all finite $\ B\subseteq A,\ $ under the canonical projections. (One could also use som other similar arguments). This inverse limit nature of $\ \mathbf S_A\ $ shows its covering 1-dimensionality:

$$\dim \mathbf(S_A)\ =\ 1$$