Choice is not needed.


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Let $X$ be a non-degenerate *continuum* (connected compact Hausdorff space). 

For a contradiction suppose $X$ is countable.  Apparently $X$ must be infinite, and so we may enumerate $X=\{x_i:i<\omega\}$  where the $x_i$'s are distinct. 

Let $U$ be a open set with $x_0\in U$ and $\overline U\neq X$.  

Let $C_0$ be a connected component of $X\setminus U$ intersecting $X\setminus \overline U$. Then $C_0$ is a non-denenerate continuum. This is true because compactness and normality of $X$ implies the quasi-component of $C_0$ is connected, and this quasi-component must meet $\partial U$ in order for $X$ to be connected.

Assuming non-degenerate continua $C_0\supseteq C_1\supseteq ... \supseteq C_{n-1}$ have been defined, let  $x^*$ be the element of $C_{n-1}$ with least subscript. 

Let $U$ be an open set with $x^*\in U$ and $C_{n-1}\setminus \overline U\neq\varnothing$. 

Let $C_n$ be a connected component of $C_{n-1}\setminus U$ intersecting   $C_{n-1}\setminus \overline U$.

Continuing in this manner, we construct a nested sequence $(C_n)$ of non-empty compact sets.  Their intersection must be non-empty.  But on the other hand we ensured each point of $X$ is eventually not in $C_n$. Contradiction.


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By contrast, it is *not* provable in ZF that every connected subset of the plane is equinumerable with the reals.  

Is every **compact** connected subset of the plane equinumerable with the reals, in ZF?  This I don't know.