Is there an improper subset that isn't equal to its superset?

Question

Can there be a set $A$ and a set $B$ such that $A\subseteq B$ and $A\ne B$ ?

While trying to find a solution to this question, I've found this answer which states:

An improper subset (usually denoted as $A\subseteq B$) is such that $A=B$ is allowed (but not mandated)

If that's true, can you give an example for such a case where $A\ne B$ ?

Answer 1

Of course, $A= \{1\}$, $B =\{1,2\}$, will do. Or $A = \emptyset$ and $B$ any non-empty set, like $\{\emptyset\}$.

Answer 2

If you can prove for every $x$ in A is also in B but inverse is not true, you will finish. This is your proving motivation. For example: let $A={(1,2,3)}$ and $B={(1,2,3,4)}$ two sets. We have for all $x$ in $A$, $x$ is also in B. But there exist a $y$ in $B$ (which is $4$) which is not in $A$. So we have finally:

$A⊆B$ and $A≠B$