>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 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:

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 $\ 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.