Proof that the relation isn't an equivalence relation

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Let $X$ be an uncountable set. We define on the set $P(X)$ a relation $S$ :

$$S = \{ (A,B) | A \in P(X) \land B \in P(X) \land A\backslash B \space \text{is a countable set} \}$$

Is $S$ reflexive, transitive, symmetric or/and antisymmetric?

I want to explain my reasoning to see if I'm correct.

Reflexivity:
$A\backslash A = \emptyset$, so $(A,A) \in S$.

Symmetry:
Assume $(A,B) \in S$, then $A\backslash B$ is countable. For the relation to be symmetric, $B \backslash A$ should be countable too. However, I give a counterexample. Suppose $A$ is countable and $B$ is uncountable, than $B\backslash A$ is uncountable. So $S$ is not symmetric.

Transitivity:
Assume $(A,B) \in S$ and $(B,C) \in S$ then $(A,C) \in S$. I have trouble finding a proof or a counterexample.

Anti-symmetric:
No, assume $A,B$ are two countable sets but with different elements, then $(B,A) \in S$ and $(A,B) \in S$ but $A \neq B$

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Symmetry : No
OP Counter-Example works.

Anti-symmetric : No
OP Counter-Example works.

Reflexivity : Yes
OP logic works.

Transitivity :
(A) When $A$ is countable , then , no matter what $B$ & $C$ & $B-C$ are , we must have countable $A-B$ & countable $A-C$
In that Case , YES
(B) When $A$ is uncountable , then $A-B$ being countable implies that $B$ has removed all those uncountable elements from $A$ , hence $B$ must be uncountable too.
Like-wise , $B-C$ being countable implies that $C$ has removed all those uncountable elements from $B$ , which are in $A$ too. Hence , when we check $A-C$ , then $C$ will remove all those uncountable elements from $A$ too.
In that Case , YES

PICTORIAL VIEW :

ABCC

When $A-B$ (Green Part) is countable , then $B$ must have removed the uncountable elements in the Inter-Section (Blue Part)
To make $B-C$ countable , $C$ can not be the Dotted Part on the right , because that will retain the uncountable elements (Blue Part).
Hence , we see that $C$ must be something like the Dotted Part in the Center , which will remove those uncountable elements from $B$ to give $B-C$.
Naturally , those Common uncountable elements will be eliminated in $A-C$ too.

It is immaterial whether $B$ had other uncountable elements in $B-A$ (Purple Part) which were removed by $C$ to give countable $B-C$.