$A \in B$ vs. $A \subset B$ for proofs

The implication $(A\in B) \wedge (B\in C) \implies A\in C$ is false. Just take $B=\{A\}$ and $C=\{B\} = \{\{A\}\}$ to have a counterexample.

Take $A=\varnothing$, $B=\{\varnothing\}$ and $C=\{\{\varnothing\}\}$.

Then clearly $A\in B\wedge B\in C$, but $A\in C$ implies $\varnothing=\{\varnothing\}$ wich cannot be true.

This because $\{\varnothing\}$ has elements and $\varnothing$ has not.

We conclude that the implication is false.

Set membership is not like putting stuff in a box.

If you put a box of chocolates into a drawer then it's certainly true that the drawer contains chocolates, but the concept of set membership is different from this. The membership only considers the direct containment, so by the mathematical language the drawer only contains a box (and in that box there's chocolates).

And that's where your reasoning is wrong (mathematically).