I have derived the proof to some extent, mentioned below:-
$$\begin{array}{rll} 1. &\lnot p \land \lnot q &\text{Premise} \\ 2. &\lnot p &\land\text{elim},1 \\ 3. &\lnot q &\land\text{elim},1 \\ 4. &p &\lnot\text{elim},2 \\ 5. &q &\lnot\text{elim},3 \\ 6. &p \rightarrow q &\rightarrow\text{intro},4,5 \\ 7. &q \rightarrow p &\rightarrow\text{intro},4,5 \\ 8. &p \Leftrightarrow q &\Leftrightarrow\text{intro},6,7 \end{array}$$
Is the above proof correct? Please correct me.
No, that is not correct.
I'm assuming it's supposed to be some kind of natural deduction system, but the deductions you annotate with $\neg$elim and $\to$into don't follow any sane negation elimination or implication introduction rules I know.
For example you try to conclude $p$ from $\neg p$. That makes no logical sense "Socrates is mortal, ergo Socrates is not mortal"??
And your introduction of the $\to$s look like you think you have a rule saying "from $p$ and $q$ conclude $p\to q$". This rule is actually sound, but it is far too weak to be useful in general, and is almost certainly not the introduction rule your text has presented.
The proof must be :
1) $\lnot p \land \lnot q$ --- premise
2) $\lnot p$ --- form 1) by $\land$-elim
3) $\lnot q$ --- form 1) by $\land$-elim
4) $p$ --- assumed [a]
5) $\bot$ --- from 2) and 4) by $\lnot$-elim
6) $q$ --- from 3) and 5) by RAA (or Double negation)
7) $p \rightarrow q$ --- from 4) and 6) by $\rightarrow$-intro, discharging [a]
8) $q$ --- assumed [b]
9) $\bot$ --- from 3) and 8) by $\lnot$-elim
10) $p$ --- from 2) and 9) by RAA (or Double negation)
11) $q \rightarrow p$ --- from 8) and 10) by $\rightarrow$-intro, discharging [b]
12) $p \leftrightarrow q$ --- from 7) and 11) by $\leftrightarrow$-intro.
