Derivative of Subset is Subset of Derivative

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

Let $A$, $B$ be subsets of $T$.

Then
 * $A \subseteq B \implies \operatorname{Der} A \subseteq \operatorname{Der} B$

where
 * $\operatorname{Der} A$ denotes the derivative of $A$.

Proof
Assume $A \subseteq B$.

Let $x \in \operatorname{Der} A$.

According to Characterization of Derivative by Open Sets it is enough to prove that for every open subset $G$ of $T$ if $x \in G$, then there exists $y$ such that $y \in B \cap G$ and $x \neq y$.

Let $G$ be an open subset of $T$.

Assume $x \in G$.

Then there exists a point $y$ of $T$ such that $y \in A \cap G$ and $x \ne y$ by Characterization of Derivative by Open Sets. $A \cap G \subseteq B \cap G$.

Hence $y \in B \cap G$ and $x \ne y$.