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 sceleton, and $\ \mathbf {0\ 1}\in S_A), is to be defined below while first (ahead of time) let's formulated
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) First an introductory observation (where $\ a\in A$):
Let $\ h\,^a\in S_A\ $ be such that $\ h\,^a\,_a:=\frac 12.\ $ Such point $\ h\,^a$ is unique, i.e. $\ p_a^{-1}(\frac 12) =\{h\, ^a\},\ $ where $\ p_a : S_A\rightarrow [0;1]\ $ is the cartesian projection.
Now the proof proper:
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\in X,\ $ i.e. $\ H := \{h\,^a\:\ a\in A\}\ \subseteq X.\ $ Thus
$$ |X|\ \ge\ |H|\ =|A|$$
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\ H:=(S\setminus\{h\})\setminus G\ $ of $\ S\ $ such that $\ a\in G\ $ and $\ b\in H$. Etc. The theorem still holds.
Logical consideration
I am not using ordered 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 to axioms like the axiom of choice or continuum hypothesis, and similar, about their relation to the cartesian product, and of the ordinary $\ [0;1]\ $ in my main example.