Derivative of Derivative is Subset of Derivative in T1 Space

Theorem
Let $T = \left({S, \tau}\right)$ be a $T_1$ topological space.

Let $A$ be a subset of $S$.

Then
 * $A'' \subseteq A'$

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

Proof
Let
 * $(1)$: $x \in A''$.

Aiming for a contradiction, suppose that $x \notin A'$.

Then by Characterization of Derivative by Open Sets there exists an open subset $G$ of $T$ such that
 * $(2)$: $x \in G$ and
 * $(3)$: $\lnot \exists y: y \in A \cap G \land x \ne y$.

By definition of $T_1$ space $\{x\}$ is closed.

Then by Open Set Less Closed Set is Open
 * $(4)$: $G \setminus \{x\}$ is open.

By $(1)$, $(2)$, and Characterization of Derivative by Open Sets there exists a point $y$ of $T$ such that
 * $(5)$: $y \in A' \cap G$ and
 * $(6)$: $x \ne y$.

Then by definition of intersection $y \in A'$.

Then
 * $(7)$: $y \in A' \setminus \{x\}$ by $(6)$ and definition of set difference.

By definition of intersection and $(5)$, $y \in G$.

$y \notin \{x\}$ by $(6)$ and definition of singleton.

Then
 * $(8)$: $y \in G \setminus \{x\}$ by definition of set difference.

We will prove
 * $(9)$: $G \cap \left({A \setminus \{x\}}\right) = \varnothing$


 * Aiming for a contradiction suppose that $G \cap \left({A \setminus \{x\}}\right) \ne \varnothing$.


 * Then by definition of the empty set there exists $g$ such that
 * $g \in G \cap \left({A \setminus \{x\}}\right)$
 * Hence by definition of intersection
 * $g \in G$ and
 * $g \in A \setminus \{x\}$.
 * Then $g \in A$ by definition of set difference.
 * Then $g \in A \cap G$ by definition of intersection.
 * Then $x = g$ by $(3)$.
 * Hence this contadicts with $g \notin \{x\}$ by definition of set difference.
 * Thus $G \cap \left({A \setminus \{x\}}\right) = \varnothing$.

Let $U = G \setminus \{x\}$ be an open set by $(4)$.

$y \in A'$ by $(5)$ and definition of set difference.

Then by $(8)$ and Characterization of Derivative by Open Sets there exists a point $q$ of $T$ such that
 * $(10)$: $q \in A \cap U$ and
 * $(11)$: $y \ne q$.

Then by definition of intersection, $q \in A$.

Then by definition of set difference
 * $(12)$: $q \in A \setminus \{y\}$.

By definition of intersection $q \in U$.

Then $q \ne x$ and $q \in A$ by $(12)$ and by definition of set difference.

Then by definition of set difference $q \in A \setminus \{x\}$

and $q \in G$.

Hence this contradicts with $(9)$ by definition of intersection.

Thus the result by Proof by Contradiction.