Choice is not needed.


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Theorem (ZF). Every non-degenerate **continuum** (connected compact Hausdorff space)  is uncountable.



*Proof*. Let $X$ be a non-degenerate continuum.

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 $C_0=X$.  

Suppose $n\geq 1$ and 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 $x^{**}$ be the element of $C_{n-1}$ with least subscript greater than $x^*$'s. Let $\mathcal U_n=\{U\in \tau:x^*\in U\text{ and }x^{**}\notin \overline U\}.$ Let $$\mathcal C_n=\{C\subseteq C_{n-1}:x^{**}\in C,\;C\text{ is connected, and }C\cap U=\varnothing\}.$$Let $C_n=\overline{\bigcup \mathcal C_n}$. Then $C_n$ is a continuum, and is non-degenerate because some elements of $\mathcal C_n$ are non-degenerate.  This is true because compactness and normality of $X$ implies the quasi-component of $x^{**}$ in $C_{n-1}\setminus U$, $U\in \mathcal U_n$, is connected, and this quasi-component must meet $\partial U$ in order for $X$ to be connected.

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. $\blacksquare$


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Follow-up question: *Is every separable metric continuum equinumerable with the reals, in ZF?*