Why $S \subset X$ disconnected in his subspace topology implies $S$ disconnected on $X$?

if $S$ are disconnected on his subspace topology then there exists $U \cap S$ where $U$ are open on $X$ and $H\cup S$ where $H$ are open in $X$ that $S=(U\cap S)\cup (H\cap S) = (H\cap U)\cup S$ but by this how can i find the open sets in $X$ such that $S$ is his union?

• You also need to add that $\emptyset = (U \cap S) \cap (H \cap S) = (U \cap H) \cap S$ of course, and $U \cap S \neq \emptyset$ and $H \cap S \neq \emptyset$. – Henno Brandsma May 11 '17 at 18:15
• You cannot find two open subsets of $X$ that exactly union up to $S$ in general. That would imply $S$ is open, in $X$ as a union of sets open in $X$, you cannot do better than $S \subseteq U \cup U$. – Henno Brandsma May 12 '17 at 4:29

$S$ disconnected on $X$ means $S$ is disconnected in its subspace topology by defination! Your question makes no sense
• S disconnected in X $\Longleftrightarrow$ S disconnected in his subspce topology, i want the details about $\Leftarrow$ and u mean $\Rightarrow$ – Eduardo Silva May 11 '17 at 19:15
• @EduardoSilva What $\Leftarrow$, $\Rightarrow$ Please state equivalence between what conditions..? – Henno Brandsma May 11 '17 at 19:26
$S = (H\cup U)\cap S$ implies $S \subseteq H \cup U$. You seem to be implying there are two definitions of a disconnected subspace, but there is only one: $S$ is disconnected in its subspace topology.