Find the mistake on the "proof" that $$\exists x(\neg p(x)\land\neg q(x))\implies\exists x\,\neg p(x)\land\exists x\,\neg q(x).$$
"Proof":
$$ \begin{array}{lll} 1)&\exists x(\neg p(x)\land\neg q(x))&\text{Premise}\\ 2)&\neg p(a)\wedge\neg q(a)&\text{Existential particularization 1)}\\ 3)&\neg p(a)&\text{Conjunction elimination 2)}\\ 4)&\neg q(a)&\text{Conjunction elimination 2)}\\ 5)&\exists x\,\neg p(x)&\text{Existential generalization 3)}\\ 6)&\exists x\,\neg q(x)&\text{Existential generalization 4)}\\ 7)&\exists x\,\neg p(x)\land\exists x\,\neg q(x)&\text{Conjunction introduction 5,6)}\\ \end{array} $$
This is not a valid proof because of line $6)$: we must not suppose that when doing the generalization, we call this $x$ but another element, namely $y$, with the possibility that $x\neq y$.
So line $6)$ should be $\exists y\,\neg q(y)$, and then we cannot apply line $7)$. Thus this is not a valid proof.
Is my reasoning correct?
No. Not correct. Just because you use the same variable twice does not mean that they have to refer to the same object. A statement like $\exists x \ Even(x) \land \exists x \ Odd(x))$ is true for the natural numbers, as indeed "there is at least one even number and there exists at least one odd number", but clearly under this interpretation the $x$'s have to refer to different objects.