Prove $ A \vdash \neg \neg A $ (Natural Deduction) From this set of rules:
$ (\wedge E) $ $A \wedge B \vdash A $
$ (\wedge I) $ $A, B \vdash A \wedge B $
$ (\vee E) $ $ A \vee B, A \rightarrow C, B \rightarrow C \vdash C $
$ (\vee I) $ $ A \vdash A \vee B $
$ (\neg E) $ $ \neg A \rightarrow B, \neg A \rightarrow \neg B \vdash A $
$ (\neg I) $ $ A \rightarrow B, A \rightarrow \neg B \vdash \neg A $
$ (\rightarrow E) $ $ A, A \rightarrow B \vdash B $
$ (\rightarrow I) $ $ Premises \vdash A \rightarrow B $
With the last rule ($ \rightarrow I $), you can introduce any implication you like as long as you prove in a sub-proof the consequent of the implication under the assumption of the hypothesis of the implication e.g.

*

*$ P \rightarrow Q \quad $ Premise

*$ \neg Q \quad\quad\space\space $ Premise

*$ P \rightarrow \neg Q \quad \rightarrow I,$ subproof below
3.1 $ P \quad\space\space\space $ Assumption
3.2 $ \neg Q \quad $ from (2)

*$ \neg P \quad $ from (1)(3), $ \neg I $
Prove $ A \vdash \neg \neg A $

*

*$ A \quad Premise $

*$ \neg (\neg \neg A) \rightarrow A \quad \rightarrow I, subproof $
2.1 $ \neg (\neg \neg A) \quad Assumption $
2.2 $ A \quad\quad\quad (1) $

*$ \neg (\neg \neg A) \rightarrow \neg (\neg \neg A) \quad \rightarrow I, subproof $
3.1 $ \neg (\neg \neg A) \quad Assumption $
3.2 $ \neg (\neg \neg A) \quad (3.1) $

*$ \neg \neg A \quad (2)(3), \neg E $
This is the proof I have currently and I'm unsure about it because of line 3 (specificially, the consequent of the implication which would usually just be $\neg A$) and subsequently line 4 which uses line 3 as a premise.
Is my proof correct?
 A: No. As you pointed out yourself, the $\lnot E$ rule requires two premises of the form $\lnot P\rightarrow Q$ and $\lnot P\rightarrow \lnot Q$. Your pair of premises $\lnot (\lnot \lnot A)\rightarrow A$ and $\lnot (\lnot \lnot A)\rightarrow \lnot (\lnot \lnot A)$ do not have this form: $Q$ would have to be both $A$ and $\lnot \lnot A$.
However, a correct proof can be given with essentially the same form as yours. The idea is simple.  We want to prove $\lnot \lnot A$. This sentence begins with $\lnot$, so we can try to prove it using the $\lnot$ introduction rule - this is a bit more natural than working with $\lnot \lnot \lnot A$ and trying to use $\lnot$ elimination! To apply $(\lnot I)$, we need to prove $\lnot A\rightarrow B$ and $\lnot A \rightarrow \lnot B$ for some choice of $B$. Can you see a $B$ that works?
Complete proof hidden in the spoiler block below.

 \begin{align*}(1) &\quad A &\quad \text{Premise}\\(2) &\quad \lnot A\rightarrow A &\quad (\rightarrow I)\\&\quad (2.1) \quad \lnot A &\quad \text{Assumption}\\&\quad (2.2) \quad  A &\quad \text{from }(1)\\(3) &\quad \lnot A \rightarrow \lnot A&\quad (\rightarrow I)\\&\quad (2.1) \quad \lnot A &\quad \text{Assumption}\\&\quad (2.2) \quad \lnot A &\quad \text{from }(2.1)\\(4) &\quad \lnot \lnot A &\quad \text{from }(2), (3), (\lnot I)\end{align*}

A: Assume $A$ and $\neg A.$ From $A$ and $\neg A$ you get $\bot$ by ($\rightarrow$-E). Thus $\neg A \rightarrow \bot$ by ($\rightarrow$-I), i.e. $\neg\neg A.$
As a diagram:
$$
\dfrac
{
  \dfrac
  {
    A
    \quad
    [\neg A]
  }
  {
    \neg
  }
  (\rightarrow\text{-E})
}
{
  \neg\neg A
}
(\rightarrow\text{-I})
$$
A: (Posted after another answer was accepted)
Using a form of natural deduction...

