For every non-trivial language $A$ and every finite strict subset $B \subsetneq A$, it's holds that $A \le_m A \setminus B$ It's claim 1 from Bader Abu Radi's solution to this question.
My solution (have no idea how wrong it is):
$B$ finite $\Rightarrow$ $B\in R \Rightarrow$ exists TM $\langle M_B\rangle$ that halts $B$.
*Wrong from here
Let's define reduction $f$ from $A$ to $A \setminus B$, in the following way $f(\langle M_B\rangle , x)=\langle M_A\rangle$
When $M_A$ implemented on input $w$ like this:

Run $M_B$ on $w$ and answer on the same way

$x\in B \Rightarrow M_B$ accept $x\Rightarrow M_A$ accept $x\Rightarrow x\in M_A$
$x\notin B \Rightarrow M_B$ reject $x\Rightarrow M_A$ reject $x\Rightarrow x\notin M_A$
till here
So the reduction $A \le_m A \setminus B$ true.
As I wrote early not sure how wrong is it, additionaly what're the changes that should be done to proof the second claim ($A \setminus B \le_m A$) from the same answer.
Edit
$f(x)=x'=x\setminus x_B$ (when $x_B$ is all the $x$ that accepted by $M_B$), because $M_B$ halts on $B$ the reduction is applicable and work for any $x$.
It's implemented in the following way:

run $M_B$ on $x$ (halts), if $M_B$ accept, $x'$ reject
otherwise return $x$

If $x\notin A \Rightarrow x'\notin A \setminus B$
If $x\in A$:
$x\in B \Rightarrow M_B$ accept $x \Rightarrow x'$ reject $\Rightarrow x'\notin A\setminus B$
$x\notin B \Rightarrow M_B$ reject $x \Rightarrow x'=x\Rightarrow x'\in A\setminus B$
Is this one correct? Should something be added to it?
As mentioned in the above comment, the format of your reduction is not correct, so you have to take care of that first, and think about it again. Also, the hint I gave in the question that you linked should be enough. Anyway, here is another hint.
Hint: consider the identity reduction that maps every word $x$ to itself, and try to understand why it does not work. Then, think what is the simplest thing you can do in order to fix the identity reduction.