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

Theorem
Let $T = \left({S, \tau}\right)$ 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 $\varnothing$.

Proof
Suppose $H \subseteq S$ were to be a non-empty subset whose boundary is empty, that is:


 * $\partial H = H^- \cap \left({S \setminus H}\right)^- = \varnothing$

Then, $H^-$ and $\left({S \setminus H}\right)^-$ are two closed sets of $T$ whose union is $S$.

Hence, by hypothesis, one of them must be empty.

By assumption, $H$ is not empty

It must therefore follow that $S \setminus H = \varnothing$.

Therefore $H = S$.

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