Can an infinite set A be larger than an infinite set B but still have the same cardinality?

Any infinite set $B$ has at least a proper subset $C$ such that $|C|=|B|$ (assuming choice, of course, or some weaker axiom thereof).

Since $B$ is infinite, it is not empty. Let $b_0\in B$ and consider $C=B\setminus\{b_0\}$.

Then $|B|=|C|$. Indeed, take a countable subset $Z$ of $B$ (it exists by choice). Then $Z\cup\{b_0\}$ is countable as well, so we can assume $b_0\in Z$. There exists a bijection $f\colon\mathbb{N}\to Z$ such that $f(0)=b_0$. Now consider $F\colon B\to C$ defined by $$ F(x)=\begin{cases} f(n+1) & \text{if $x\in Z$ and $x=f(n)$} \\[4px] x & \text{if $x\notin Z$} \end{cases} $$ It's easy to prove that $F$ is a bijection.

If $|A|=|B|$, then we can use $F$ to provide also a bijection $A\to C$ and $C$ is a proper subset of $B$.