Let $A$ be any uncountable set, and let $B$ be a countable subset of $A$. Prove that the cardinality of $A = A - B $ I am going over my professors answer to the following problem and to be honest I am quite confused :/ Help would be greatly appreciated!

Let $A$ be any uncountable set, and let $B$ be a countable subset of $A$. Prove that $|A| = |A - B|$ 

The answer key that I am reading right now follows this idea:
It says that $A-B$ is infinite and proceeds to define a new denumerable subset $A-B$ as $C$. Of course since $C$ is countably infinite then we can write $C$ as ${c1,c2,c3...}$ 
Once we have a set $C$, we know that the union of $C$ and $B$ must be denumerable (from another proof) since $B$ is countable and $C$ is denumerable.
This is where I start to have trouble. The rest of the solution goes like this...
Since the union of $C$ and $B$ is denumerable, there is a bijective function $f$ that maps the union of $C$ and $B$ to $C$ again. The solution then proceeds to define another function $h$ that maps $A$ to $A-B$. 
I am just so lost. The thing is I don't even understand the point of constructing a new subset $C$ or defining functions like $f$ or $h$.
So I suppose my question is in general, how would one approach this problem? I am not mathematically inclined unfortunately, and a lot of the steps in almost all of these problems seems arbitrary and random. Help would be really appreciated on this problem and some general ideas on how to solve problems like these!!!
Thank you so very much! 

Your answer key is right, but I think it can be describe more simply.
We have $A$ uncountable and $B$ countable.  Since $A-B$ is infinite, it contain another countable set $C$.  So there is a bijection $C\rightarrow B\cup C$ as you note.  Now let $D=(A-B)-C$, so $A-B=C\cup D$, and therefore $A=B\cup C\cup D$, all three of which are disjoint from one another. Since you have the bijection $C\rightarrow B\cup C$, you clearly have another bijection $(C)\cup D\rightarrow (B\cup C)\cup D$, but by construction that is the bijection $A-B\rightarrow A$.
But why is it done this way?  It's kind of a trick.  The key is realizing that you needed to separate both $A$ and $A-B$ into a countable piece, and another piece whose cardinality may not be countable.  Further, you need that questionable piece to be the same set on both sides of the bijection: that's $D$ in my example.  The trick is that the countable piece is different on each side of the bijection ($C$ vs. $B\cup C$), but since both are countable you get a bijection anyway!
