Equivalence of Definitions of Connected Topological Space/No Union of Closed Sets implies No Subsets with Empty Boundary

Theorem
Let $T = \struct {S, \tau}$ be a topological space.

Let $T$ have no two disjoint non-empty closed sets whose union is $S$.

Then the only subsets of $S$ whose boundary is empty are $S$ and $\O$.

Proof
Let $H \subseteq S$ be a non-empty subset whose boundary $\partial H$ is empty.

Thus:

From Topological Closure is Closed, both $H^-$ and $\paren {S \setminus H}^-$ are closed sets of $T$.

From Union of Closure with Closure of Complement is Whole Space:
 * $H^- \cup \paren {S \setminus H}^- = S$

Thus $H^-$ and $\paren {S \setminus H}^-$ are two disjoint closed sets of $T$ whose union is $S$.

Hence,, one of them must be empty.

Suppose $H$ is not empty.

It must therefore follow that:
 * $S \setminus H = \O$

Therefore $H = S$.

Thus the only subsets of $S$ whose boundary is empty are $S$ and $\O$.