Is this a right proof proving that if $A\rightarrow False$ is true then A is logically equivalent to False? Here's my proof:
$$\left(A\rightarrow \bot\right)\rightarrow\left( A\rightarrow \bot \wedge \bot\rightarrow A\right)$$
$$\left( A\rightarrow \bot \wedge \bot\rightarrow A\right)\rightarrow\left(A\equiv \bot\right)$$
both of statements are true so it proves that if $A\rightarrow \bot$ is true then $A\equiv \bot$ is also true which should mean A is logically equivalent to $\bot$.

The proof is not correct because there is no rule saying that $A\to B \to((A\to B)\wedge(B\to A))$. On the other hand, the definition of $\to$ is
that $A\to B \equiv (\lnot A)\vee B$, so that
$A\to False \equiv (\lnot A)\vee False \equiv \lnot A$
You should be able to conclude.
Use existing rules, don't use rules that don't exist!
Edit Wikipedia calls the rule $A\to B \equiv (\lnot A)\vee B$ material implication. It is often the easiest way to handle $\to$ because the semantics of negation and the disjunction connector are straightforward.
Edit 2 Following @AsafKaragila's comments, it occurs to me that the OP's proof may have used implicitly that $False \to A \equiv True$ so that $(A\to False) \wedge (False\to A)$ is equivalent to $A\to False$.
This may validate the original proof, but I still believe the proof with material implication is better.
