Is $A$ \ $B$ ~ $A$ true? 1) If $A$ is infinite and uncountable, and $B$ is finite or countable, is $A$ \ $B$ ~ $A$ true?

2) If $A$ is infinite uncountable and $B \subset A$ is finite or countable, is $A$ \ $B$ ~ $A$ true?

1)I assume it is true, because of the cardinality of the continuum?
2)Also true for the same reason?
1) follows from 2), because in all cases $A \setminus B = A \setminus (A \cap B)$, and if $B$ is finite or countably infinite, so is $A \cap B$.

To prove 2), if $B$ is finite or countable, it is not all of $A$. So there exists $x \in A \setminus B$. Now you can apply the same argument to $A$ and $B \cup \{x\}$. Repeat. You will be able to find a countable subset $C$ of $A$, disjoint from $B$. Now: $A \setminus B = (A \setminus (B \cup C))\cup C$ and $A = (A \setminus(B\cup C)) \cup (B \cup C)$. The "outer" unions are disjoint unions in both cases.

Now compare the RHS of the two equations: The first term of the union is the same, and the second is a countably infinite set in both cases. There is a bijection between $C$ and $B\cup C$, and you can extend this with the identity map on $A\setminus(B\cup C)$ to get a bijection from $A\setminus B$ to $A$.
They are indeed both true. 
Assume that $A$ is uncountable, $B$ is countable, and $A - B$ is countable. Since the union of two countable sets is countable, then $(A - B) \cup B$ is countable. But $A \subset (A - B) \cup A;$ since a subset of a countable set is itself countable, we conclude that $A$ has to be countable. This contradiction proves that $A - B$ is uncountable.