Closure of Subset of Closed Set of Topological Space is Subset/Proof 2

Proof
Let $x \notin F$.

Then $x$ is in the open set $S \setminus F$.

From Set Difference with Subset is Superset of Set Difference:
 * $S \setminus F \subseteq S \setminus H$

From Subsets of Disjoint Sets are Disjoint:
 * $S \setminus F \cap H = \O$

From Set is Open iff Neighborhood of all its Points:
 * $S \setminus F$ is an neighborhood of $x$

By definition of the closure of $H$:
 * $x \notin H^-$

We have shown that:
 * $S \setminus F \subseteq S \setminus H^-$

From Set Difference with Subset is Superset of Set Difference:
 * $H^- \subseteq F$