Let $X=\mathcal{C}[0,1]$, $A=\{x \in X:$ the set $x \enspace [0,1] \cap (\mathbb{R\setminus \mathbb{Q}})$ is at most countable $\}$. Is the set $A$ connected in $X$?
A set is at most countable iff it is either finite or countable infinite. I think that the set A is disconnected by the following argument. By using the definition of disconnected sets.
$A \subset (X,d)$ is disconnected if $A= A_1 \cup A_2$ with $A_k \neq \emptyset, A_1, A_2$ being separated sets.
Take $A_1=B(x_1,\varepsilon_0/3)$, $A_2 = \bigcup^{\infty}_{n=2} B(x_n,\varepsilon_0/3 )$ such that $A \subset A_1 \cup A_2$ being separated such that $\bar{A_1} \cap A_2 = \emptyset$. A is disconnected.
Is my thinking correct? Any help would be apppreciated.
$A$ is precisley the set of all constant functions since a continuous function with a countable range is a constant (by IVP). Hence $A$ is connected: any convex set in a normed linear space is connected.