Suppose that $A$ and $B$ are two infinite sets and $|A|<|B|$. The question is that how to prove that $|A∪B|=|B|$. The proof is related to the Axiom of Choice.
$|A\cup B| \leq |A| + |B| \leq |B| + |B| = |B|$ since $|B|$ is of infinite cardinality and $2\cdot \aleph_k = \aleph _k$ for whichever cardinal $|B|$ happens to be.
Then also $|B|\leq |A\cup B|$ by subadditivity.
Hence $|A\cup B| = |B|$ when $B$ is an infinite set and $|A|\leq |B|$
I don't think you need to use the axiom of choice for the proof, though perhaps I have naive notions on how set operations work on higher sized sets.
By the axiom of choice all infinite cardinals are alephs, so you can use $n+\aleph_0=\aleph_0$ for finite $|A|=n$, and $2\cdot\aleph_\alpha=\aleph_\alpha$ for infinite $|A|=\aleph_\alpha$.
In the absence of the axiom of choice, there could be an infinite cardinal $m$ such that $m\lt2m$. From $m\lt2m$ it follows (this is not trivial) that $2m\lt3m$. Then, if $A,B$ are disjoint sets with $|A|=m$ and $|B|=2m$, we have $|A|\lt|B|$ while $|A\cup B|=m+2m=3m\gt2m=|B|$.