I don't think choice is needed.
For a contradiction suppose $X=\{x_i:i<\omega\}$ is a countable continuum (connected compact Hausdorff), 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.
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.