Q.Let $A = \{1, 2\}$ and $B = \{1, 2, 3\}$. Verify that $A \subset B \Rightarrow A \subseteq B$. Then, using $C = \{1, 2\}$ and $D = \{1, 2\}$, show that the converse — $A \subseteq B \Rightarrow A \subset B$ — is not always true.

First confirm the proper-subset case also satisfies the plain-subset definition; then find a case where the plain-subset condition holds but the sets are equal, so "proper" fails.

Part 1 — $A \subset B \Rightarrow A \subseteq B$ for $A=\{1,2\}$, $B=\{1,2,3\}$.

Every element of $A$ (namely $1$ and $2$) is in $B$, so $A \subseteq B$ holds. Also, $B$ has the extra element $3$ that $A$ lacks, so $A \ne B$; combined with $A \subseteq B$, this gives $A \subset B$. Since $A \subset B$ is true and we've just shown $A \subseteq B$ is also true, this confirms $A \subset B \Rightarrow A \subseteq B$ in this case — as it must always be, since "proper subset" is defined as "subset, and not equal," so it can never fail the plain subset condition.

Part 2 — Testing the converse with $C = \{1,2\}$, $D = \{1,2\}$.

Every element of $C$ is in $D$ (they are the same two elements), so $C \subseteq D$ holds. But $C = D$ (both are exactly $\{1,2\}$), so $C$ is not a proper subset of $D$ — $C \subset D$ is false. …