If A is an infinite set and B is at most countable set, prove that A and $A \cup B$ have the same cardinality I am trying to prove:
If A is an infinite set and B is at most countable set, prove that A and $A \cup B$ have the same cardinality.
Set definitions here are from baby Rudin. 
Attempt:
There are 4 possible cases:
1.A is countable and B is finite.
2.A is countable and B is countable.
3.A is uncountable and B is finite.
4.A is uncountable and B is countable.
Then for each case I need to show that $A \cup B \sim A$. I can prove 1-3, but I am unsure how to prove case 4. And in general, it seems that my proof strategy is inefficient and there exist much faster proof then proof by cases.
 A: First prove that the union of two at most countable sets is at most countable.
Now show that any infinite set has a countably infinite subset.
Do you see how to combine these facts to solve the problem?
A: Define a set $S$ to be finite iff there is a bijection from $S$ to $\{j\in \mathbb N_0:j<n\}$ for some $n\in \mathbb N_0=\mathbb N \cup \{0\}.$ It is a consequence of the Axiom of Choice (AC) that an infinite set has a countably infinite subset.
Assume (AC). Let $A$ be infinite. Let $T\subset A$ where $T$ is countably infinite. Let $B$ be countable. Let $B'=B$ \ $A.$ Then $B'$ is countable and $A\cup B=A\cup B'$ and $A \cap B'=\emptyset.$ 
Now $T$ and $T\cup B'$ are countably infinite. So there is a bijection $f:T\cup B'\to T.$ Extend $f$ to the domain $A\cup B'=A\cup B$ by letting $f(x)=x$ for $x\in A$ \ $T.$ Then $f:A\cup B\to A$ is a bijection.
A: Since $A$ is infinite, there exists $C\subset A$ countable, since $B$ is at most countable, $C\bigcup B$ as the cardinality of $C$ so the cardinality of $A=(A-C)\bigcup C$ is the cardinality of $(A-C)\bigcup (C\bigcup B)=A\bigcup B$.
A: In case 4, pick any countable subset $A' \subset A$. Then $A \cup B = (A \smallsetminus A') \cup (A' \cup B)$. Now apply case 2.
