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! 

$|A-B|\leq |A|$ is obvious.
For the reverse inequality, 
Use $|A|=|A\times A|$. 
Denote the bijection $A\rightarrow A\times A$ by $\phi$, then $\phi(B)$ embeds into a countable subset of $A\times A$. 
Then consider the projection $\pi_1$ onto the first coordinate, of the set $\phi(B)$. 
Namely $\pi_1 \phi(B)$. 
Since $|\pi_1\phi(B)|\leq |B|$, and $|B|<|A|$, we see that $A-\pi_1\phi(B)$ is nonempty. Pick any element $x \in A-\pi_1\phi(B)$, then we see that 
$\{x\}\times A$ embeds into $(A\times A)-\phi(B)$. 
This gives the reverse inequality $|A|\leq |A-B|$. 
