This is a proof with Natural Deduction, using only intuitionistically valid rules, that Double Negation implies Peirce's law :
1) $\lnot \lnot p \rightarrow p$ --- assumed [a]
2) $(p \rightarrow q) \rightarrow p$ --- assumed [b]
3) $\lnot p$ --- assumed [c]
4) $p$ --- assumed [d]
5) $\bot$ --- from 3) and 4) by $\rightarrow$-elimination
6) $q$ --- from 5) by $\bot$-rule : from $\bot$, infer $\varphi$
7) $p \rightarrow q$ --- from 4) and 6) by $\rightarrow$-introduction, discharging [d]
8) $p$ --- from 7) and 2) by $\rightarrow$-elimination
9) $\bot$ --- from 3) and 8) by $\rightarrow$-elimination
10) $\lnot \lnot p$ --- from 3) and 9) by $\lnot$-introduction, discharging [c]
11) $p$ --- from 1) and 10) by $\rightarrow$-elimination
The proof, up to now, amounts to : $\lnot \lnot p \rightarrow p, (p \rightarrow q) \rightarrow p \vdash p$
12) $((p \rightarrow q) \rightarrow p) \rightarrow p$ --- from 2) and 11) by $\rightarrow$-introduction, discharging [b], i.e. : $\lnot \lnot p \rightarrow p \vdash ((p \rightarrow q) \rightarrow p) \rightarrow p$
13) $\vdash (\lnot \lnot p \rightarrow p) \rightarrow (((p \rightarrow q) \rightarrow p) \rightarrow p)$ --- from 12) by $\rightarrow$-introduction.
Note : with classical logic, we can avoid the premise : $(\lnot \lnot p \rightarrow p)$ and prove : $\vdash ((p \rightarrow q) \rightarrow p) \rightarrow p$ using Double Negation (which is not intuitionistically valid) in step 11).
Hint
1) $∃xP(x) ∧ ∃x¬P(x)$ --- premise
2) $∃xP(x)$ --- $∧$-elim
3) $∃x¬P(x)$ --- $∧$-elim
4) $P(a)$ --- assumed [a] from 2) for $∃$-elim
5) $¬P(b)$ --- assumed [b] from 3) for $∃$-elim
6) $\forall x \forall y (x=y)$ --- assumed [c]
7) $a=b$ --- from 6)
8) $\bot$ --- from 4), 5), 7) and $=$-elim.
Now we can discharge [a] and [b] by $∃$-elim from 2) and 3) and apply $\to$-intro with 6) to conclude with :
$¬\forall x \forall y (x=y)$.
Now, with Double Negation, we may derive : $∃x ∃y \ \lnot (x = y)$.
Best Answer
Under the premise $\neg\exists x P(x)$, assume $P(x)$ and follow from this contradiction.