Set is Closed iff it Contains its Boundary

Theorem
Let $T$ be a topological space, and let $H \subseteq T$.

Then $H$ is closed in $T$ :
 * $\partial H \subseteq H$

where $\partial H$ is the boundary of $H$.

Proof
From Boundary is Intersection of Closure with Closure of Complement:
 * $\partial H = H^- \cap \left({T \setminus H}\right)^-$

where $H^-$ is the closure of $H$.

Hence from Intersection is Subset we have that:
 * $\partial H \subseteq H^-$

Then from Closed Set Equals its Closure, $H$ is closed in $T$ $H = H^-$.

Hence the result.