So Q(x) is a predicate in some domain of discourse. I am given $\neg$($\forall$y Q(y)) and I want to make a formal proof that shows that $\exists$x (Q(x) $\rightarrow$ $\forall$y Q(y)) follows from it. How would one approach it?
I tried moving the negation all the way to the center with De Morgans. Then I did Elim $\exists$ for some special c that satisfies Q. Then I did Intro $\exists$ to introduce x. I am not sure where to go from there.
You say:
Then I did Elim $\exists$ for some special c that satisfies Q.
That's not right. If you moved the negation inside, you should have gotten $\exists y\ \color{red}{\neg} Q(y)$. So, your special $c$ should not have property $Q$.
In fact, with $\neg Q(c)$, you should be able to to prove $Q(c) \to \forall y \ Q(y)$, because the antecedent of that conditional is false, and so the conditional has to be true. So by $\exists$ Intro, you're there.
If you want help with the xact formalization of that, you'll need to let us know how exactly your rules are formally defined, since there are many different formalization. But, I think with the above proof sketch, you should be able to do it.
$\neg \forall y Q(y)$
$\exists y \neg Q(y)$ by Demorgan's Law for Quantifers
$\neg Q(a)$ by existential instantiation
$\neg Q(a) \vee Q(y)$ by disjunction introduction
$Q(a) \rightarrow Q(y)$ by implication rule
$\forall y (Q(a) \rightarrow Q(y))$ by universal generalization
$Q(a) \rightarrow \forall y Q(y)$ by null quantification
$\exists x (Q(x) \rightarrow \forall y Q(y))$ by existential generalization
