Element Depends on Independent Set iff Union with Singleton is Dependent/Lemma

Theorem
Let $M = \struct{S, \mathscr I}$ be a matroid.

Let $A, X \in \mathscr I$.

Let $x \in S : x \notin X$.

Let $X \cup \set x$ be dependent.

Let $A \subseteq X \cup \set x$.

Then:
 * $\size A \le \size X$

Case 1: $x \in A$
From Set Difference over Subset:
 * $A \setminus \set x \subseteq \paren{X \cup \set x} \setminus \set x$

From Set Difference is Right Distributive over Union:
 * $\paren{X \cup \set x} \setminus \set x = \paren {X \cup \set x} \cup \paren{\set x \setminus \set x}$

From Set Difference with Disjoint Set:
 * $\paren {X \cup \set x} = X$

From Set Difference with Superset is Empty Set:
 * $\paren{\set x \setminus \set x}$

Thus:
 * $A \setminus \set x \subseteq X$

Aiming for contradiction:
 * $A \setminus \set x = X \implies X \cup \set x = \paren {A \setminus \set x} \cup \set x$

From Set Difference Union Second Set is Union:
 * $\paren {A \setminus \set x} \cup \set x$

Contradicting:
 * $X \cup \set x$ is dependent

So:
 * $A \setminus \set x \subset X$

From Cardinality of Proper Subset of Finite Set:
 * $\size {A \setminus \set x} < \size X$

From Cardinality of Set Difference with Subset:
 * $\size {A \setminus \set x} = \size A - \size {\set x}$

From Cardinality of Singleton:
 * $\size {A \setminus \set x} = \size A - 1$

So:
 * $\size A - 1 < \size X$

So:
 * $\size A \le \size X$

Case 2: $x \notin A$
Let $x \notin A$.

Then:
 * $\paren {X \cup \set x} \cap A = \paren{X \cap A} \cup \paren{\set x \cap A = \paren{X \cap A} \cup \O = X \cap A$

From :
 * $A \subseteq X$

From :
 * $\size A \le \size X$

In either case:
 * $\size A \le \size X$