Let \(B={b0,b1,b2,...}\)

If C were finite, we would have that \(\displaystyle{A}={B}∪С\) is countable, which is a contradiction. Thus C is infinite, thus it has a countable suset \(\displaystyle{D}⊂{C}\).

Let: \(D={d0,d1,d2,...}\)

Then we can define a function f: \(A \rightarrow C\) \(\displaystyle{f{{\left({x}\right)}}}={\left\lbrace{d}{2},{x}={b};{d}{2}+{1},{x}={d};{x},{x}∈{A}{\left({B}∪{D}\right)}\right.}\)