Let $U \subseteq \Bbb R^k$ be an open subset which is not connected. Then there exists a smooth locally constant surjective function $f : U \longrightarrow \{0,1\}.$
What I know is that if $U$ is not connected then there exist non-empty disjoint open subsets $A,B \subseteq U$ such that $U = A \cup B.$ So if we consider a function $f : U \longrightarrow \{0,1\}$ such that $f(A) = \{0\}$ and $f(B) = \{1\}$ then $f$ would be a locally constant continuous surjective function from $U$ onto $\{0,1\}$ and then $Df(\textbf {x})$ is zero on a neighbourhood of $0$ for any $\textbf {x} \in U.$ Since $Df(\textbf {x})$ is a linear functional defined on the normed linear space $\left (\Bbb R^k, \|\cdot\|_2 \right )$ it follows that $Df(\textbf {x}) \equiv 0$ on $\Bbb R^k$ for any $\textbf {x} \in U.$ Hence $Df : U \longrightarrow L(\Bbb R^k,\Bbb R)$ will be a zero function and hence all it's higher order derivatives exist and they are all identically zero. But that means $f \in C^{\infty} \left (U, \Bbb R \right ).$ This completes the proof.
Does this above proof make sense?