Set of Condensation Points of Union is Union of Sets of Condensation Points/Lemma

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

Let $A, B$ be subsets of $S$. Let $x$ be a point of $S$.

Then:
 * if $x$ is condensation point of $A \cup B$,
 * then $x$ is condensation point of $A$ or $x$ is condensation point of $B$.

Proof
Assume $x$ is condensation point of $A \cup B$.

Aiming for a contradiction suppose that
 * $x$ is not condensation point of $A$ and $x$ is not condensation point of $B$.

By definition of condensation point:
 * $\exists U_1 \in \tau: x \in U_1 \land A \cap U_1$ is countable

By definition of condensation point:
 * $\exists U_2 \in \tau: x \in U_2 \land B \cap U_2$ is countable

By definition of intersection:
 * $x \in U_1 \cap U_2$

By definition of topological space:
 * $U_1 \cap U_2 \in \tau$

By Intersection is Subset:
 * $A \cap U_1 \cap U_2 \subseteq A \cap U_1 \land B \cap U_1 \cap U_2 \subseteq B \cap U_2$

By Set Union Preserves Subsets:
 * $\left({A \cap U_1 \cap U_2}\right) \cup \left({B \cap U_1 \cap U_2}\right) \subseteq \left({A \cap U_1}\right) \cup \left({B \cap U_2}\right)$

By Countable Union of Countable Sets is Countable:
 * $\left({A \cap U_1}\right) \cup \left({B \cap U_2}\right)$ is countable

Then by Subset of Countable Set is Countable:
 * $(1): \quad \left({A \cap U_1 \cap U_2}\right) \cup \left({B \cap U_1 \cap U_2}\right)$ is countable

By Intersection Distributes over Union:
 * $\left({A \cap U_1 \cap U_2}\right) \cup \left({B \cap U_1 \cap U_2}\right) = \left({A \cup B}\right) \cap \left({U_1 \cap U_2}\right)$

By definition of condensation point:
 * $\left({A \cup B}\right) \cap \left({U_1 \cap U_2}\right)$ is uncountable

This contradicts $(1)$.

Thus the result fallows by Proof by Contradiction.