Closure of Dense-in-itself is Dense-in-itself in T1 Space

Theorem
Let $T$ be a $T_1$ topological space.

Let $A \subseteq T$.

Let $A$ be dense-in-itself.

Then the closure $A^-$ of $A$ is also dense-in-itself.

Proof
Let $A$ be dense-in-itself.

Then by Dense-in-itself iff Subset of Derivative:
 * $(1): \quad A \subseteq A'$

where $A'$ denotes the derivative of $A$.

By Derivative of Derivative is Subset of Derivative in $T_1$ Space:
 * $(2): \quad A'' \subseteq A'$

By Dense-in-itself iff Subset of Derivative it is sufficient to proof that:
 * $A^- \subseteq \left({A^-}\right)'$

Thus: