We say $A \subseteq \mathbb{R}^n$ is disconnected when there exists open sets $U$ and $V$ in $\mathbb{R}^n$ such that $$U \cap A \neq \emptyset,V \cap A \neq \emptyset, U \cap V=\emptyset, \text{and} A \subseteq U \cup V$$. We say $A$ is connected if it is not disconnected.
Let $A\subseteq S \subseteq \mathbb{R}^n$. Show that if $A$ is connected in $\mathbb{R}^n$ if and only if $A$ is connected in $S$.
The forward direction is simple, but I don't know how to prove the other direction.
Here's my proof: If $A$ is disconnected in $S$,I want to show that $A$ is disconnected in $\mathbb{R}^n$. then there exist sets $U,V \subseteq S$ which are open in $S$ such that $$U \cap A \neq \emptyset, V\cap A \neq \emptyset, U \cap V=\emptyset,A \subseteq U \cup V$$
Since $U,V$ are open in $S$, there exist open sets $U_0,V_0$ in $\mathbb{R}^n$ such that $U=U_{0}\cap S$ and $V=V_{0} \cap S$. We can show that $$U_0 \cap A \neq \emptyset, V_0 \cap A \neq \emptyset, U_0 \cap V_0 = \emptyset$$ But how can i show that $A \subseteq U_0 \cup V_0$ I can only show $A \subseteq S \cap (U_0 \cup V_0)$.