Choice is not needed.
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 $U_0$ be a open set with $x_0\in U_0$ and $\overline {U_0}\neq X$.
Let $C_0$ be a connected component of $X\setminus U_0$ intersecting $X\setminus \overline {U_0}$. 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 $\partial U_0$ 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 $U_n$ be an open set with $x^*\in U_n$ and $C_{n-1}\setminus \overline U_n\neq\varnothing$.
Let $C_n$ be a connected component of $C_{n-1}\setminus U_n$ intersecting $C_{n-1}\setminus \overline {U_n}$.
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. $\blacksquare$
Follow-up question: Is every separable metric continuum equinumerable with the reals, in ZF?