The assertion you want to prove is equivalent (in ZF) to the assertion that every infinite set has a countably infinite subset, an easy consequence of the axiom of choice which can't be proved in ZF.
First, suppose there is an infinite set $D$ with no countably infinite subset. Let $A=\mathbb N\cup D$ and let $B=\mathbb N.$ Then $A$ is uncountably infinite, $B$ is a countably infinite subset of $A$, but there is no injection from $B$ to $A\setminus B$ since $A\setminus B\subseteq D.$
Now, assume that every infinite set has a countably infinite subset. Suppose $A$ is an uncountably infinite set, and $B$ is a countable subset of $A.$ Then $A\setminus B$ is an infinite set (else $A$ would be countable), whence there is a countably infinite set $C\subseteq A\setminus B$. Now, writing $+$ for disjoint union, we have
$$A=[A\setminus(B\cup C)]+(B\cup C)$$
$$A\setminus B=[A\setminus(B\cup C)]+C.$$
Then $B\cup C$ is countably infinite, so there is a bijection from $B\cup C$ to $C,$ which can be extended to a bijection from $A$ to $A\setminus B.$