Choice is not needed.
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.
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.