Definition:Connected (Topology)/Set

Definition 1 implies Definition 2
Suppose that $H$ is disconnected by Definition 1. Then $H = A \cup B$, where $A$ and $B$ be nonempty separated sets of $T$. Then by Separated Sets are Clopen in Union, $A$ and $B$ are open in $H$, considered with the subspace topology. Thus there are open sets $U$ and $V$ in $S$ such that $A = U \cap H$ and $B = V \cap H$.

Then $U \cap H \ne \varnothing$, $V \cap H \ne \varnothing$,

All that remains is to show that $U \cap V \cap H = \varnothing$. But

Definition 2 implies Definition 1
Suppose $H$ is disconnected by definition 2. Then there are open sets $U$ and $V$ in $S$ such that $U \cap H \ne \varnothing$, $V \cap H \ne \varnothing$, $H \subseteq U \cup V$, and $U \cap V \cap H = \varnothing$.

Let $A = U \cap H$ and let $B = V \cap H$.

Then $A \ne \varnothing$ and $B \ne \varnothing$, by hypothesis.

Since $H \subseteq U \cup V$, $H = H \cap \left({U \cup V}\right) = \left({H \cap U}\right) \cup \left({H \cap V}\right) = A \cup B$.

It remains to show that $A$ and $B$ are separated. Since $H = A \cup B$ and $U \cap V \cap H = \varnothing$,

$A = H \setminus B$. Since $V$ is open in $S$, $B$ is open in $H$, so $A$ is closed in $H$.

Thus $A = A^- \cap H$, so (since $B\subseteq H$) $A^- \cap B \subseteq A^- \cap B \cap H = A \cap B = \left({U \cap H}\right) \cap \left({V \cap H}\right) = U \cap V \cap H =\varnothing$.

A similar argument shows that $A \cap B^- = \varnothing$, so $A$ and $B$ are separated.

Definition 2 implies Definition 3
Suppose $H$ is disconnected by Definition 2. Then there are open sets $U$ and $V$ in $S$ such that $U \cap H \ne \varnothing$, $V \cap H \ne \varnothing$, $H \subseteq U \cup V$, and $U \cap V \cap H = \varnothing$.

Let $U' = U \cap H$ and $V' = V \cap H$. By assumption, $U'$ and $V'$ are non-empty.

By the definition of the subspace topology, $U'$ and $V'$ are open in $H$.

Since $H \subseteq U \cup V$, $H = H \cap \left({U \cup V}\right) = \left({H \cap U}\right) \cup \left({H \cap V}\right) = U' \cup V'$.

Thus $U'$ and $V'$ form a separation of $H$.

Definition 3 implies Definition 2
Suppose $H$ is disconnected by Definition 3. Then there exist non-empty, disjoint open sets $U'$ and $V'$ in $H$ such that $H = U' \cup V'$.

By the definition of the subspace topology, there are open sets $U$ and $V$ in $S$ such that $U' = U \cap H$ and $V' = V \cap H$.

Then $U \cap V \cap H = \left({U \cap H}\right) \cap \left({V \cap H}\right) = U' \cap V' = \varnothing$.

Since $U'$ and $V'$ are non-empty, $U' = U\cap H$, and $V' = V \cap H$, $U \cap H \ne \varnothing$ and $V \cap H \ne \varnothing$.

Since $H = U' \cup V'$, $U' \subseteq U$, and $V' \subseteq V$, $H \subseteq U \cup V$.