Let $B$ and $C$ be two subsets of $A$ having the same cardinality and cardinality of $A$ is strictly greater than that of the sets $B$ and $C$ then can I conclude that $A \setminus B$ and $A\setminus C$ they both have the same cardinalities?
I constructed this question myself and have found no counterexample but I am also unable to prove it. Any help will the truly appreciated.
As explained in the comments, normal integer arithmetic proves this in the finite case. In the infinite case, we can use choice and the assumption that $A$ has strictly larger cardinality than $B$ or $C$ to prove it. This assumption on $A$ is critical, lest $A = B = \mathbb Z$ and $C = 2 \mathbb Z$ be a counterexample.
The application of choice here is that for two infinite cardinals $\lambda, \kappa$ we have $\lambda + \kappa = \max(\lambda, \kappa)$. We have $|A| = |A - B| + |B| = \max(|A - B|, |B|)$. As we assume $|B| < |A|$ we must therefore have $|A - B| = |A|$. Similarly, $|A - C| = |A|$, so $|A - B| = |A - C|$. We didn't even need the assumption that $|B| = |C|$.