Suppose $A\subset B$ and $A\sim(A\cup C)$. Prove that $B\sim(B\cup C)$. 
Suppose $A\subset B$ and $A\sim(A\cup C)$. Prove that $B\sim(B\cup C)$.

We can write $A\cup C=A\cup(A^c\cap C)$ and $B\cup C=B\cup(B^c\cap C)$. And we have $B^c\cap C\subset A^c\cap C$. I don't know how to continue.
And I wonder whether the following propsition is true
If $X$ is an infinite set, $|Y|\leq |X|$, then $X\cup Y\sim X$.
If the propsition is true, then I think I can complete the proof.
Apprieciate any idea or help!
 A: Your last statement requires Choice. Without choice we can have an infinite set $X$ and a set $Y$ that is smaller in cardinality than $X$ and yet $X \cup Y$ is bigger in cardinality than $X$, such sets are called Dedekind finite sets that are Tarksi infinite. Those sets are not bijective to any proper subset of them. But with Choice, all infinite sets are Dedekind infinite, and then yes you have: $$|Y| \leq |X| \implies |Y \cup X|=|X|$$, and it's easy to complete the proof of your statement given Choice.
Back to your question, I shall not assume Choice. Now if $C \subset A$ then the result is trivially true. If $C \not \subset B$, then $A$ would be Dedekind infinite (since it is equal in size to a proper subset of it), and so $B$ would be Dedekind infinite, and so the result would hold. Now to prove the result of your main question we start with some bijection $f$ from $A$ to $A \cup C$, we take a subset $f^*$ of $f$ whose range is the set $(B \cap C) \setminus A$, now $dom(f^*)$ is a subset of $A$. Now what we do is that we take the union of $dom(f^*)$ and the set $B \setminus A$, call that union as the set $D$. Now notice that $([=] \restriction D) \cup (f \setminus f^*)$ is a bijection from $B$ to $B \cup C$.
