I am trying to prove that $\neg A \to \bot \vdash A$, I can only use double negation elimination to get the conclusion, but I don't know how to prove double negation elimination.
Below is my proof for $\neg A \to \bot \vdash A$:
1 $\neg A\to\bot$
2 $\neg A$ (assumption)
3 $\bot$ ($\to$e 1, 2)
4 $\neg\neg A$ ($\neg$i 2-3)
5 $A$ (DNE 4)
The rules that can be used would be the basic rules of natural deduction: 
The rules you list comprise a natural deduction system for intuitionistic logic. The sequent $\lnot A\to \bot\vdash A$ is not valid in intuitionistic logic, so it is not provable in your system.
To prove $\lnot A\to \bot\vdash A$, you need to be working with a natural deduction system for classical logic, which means adding an additional rule to your system. One common choice is to add double negation elimination itself; this makes your proof a valid one. Other common choices are adding reductio ad absurdum (proof by contradiction) or the law of excluded middle.
