Natural deduction proof of $\forall x (\exists y (P(x) \vee Q(y))) \vdash \exists y (\forall x (P(x) \vee Q(y)))$

You can't go from step 3 to step 4. For the particular $v$ you chose, you know that you can find $w$ such that $P(v)\vee Q(w)$. If $Q(w)$ is true, then certainly $P(x)\vee Q(w)$ is true for all $x$, but if $Q(w)$ is false there's no reason why it should be the case that $P(x)\vee Q(w)$ for all $x$.

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I don't know how to write this in the Fitch format, but the essential idea is just that either $Q(y)$ is true for some $y$, or $Q(y)$ is false for all $y$.

If $Q(v)$ is true for some particular $v$, then $P(x)\vee Q(v)$ is true for all $x$, and so we certainly have that $\exists y(\forall x(P(x)\vee Q(y)))$; the $y$ that exists is just $y=v$.

If $Q(y)$ is false for all $y$, then $P(x)\vee Q(y)\iff P(x)$, so $\forall x(\exists y(P(x)\vee Q(y)))$ implies that $P(x)$ is true for all $x$, and hence $\exists y(\forall x(P(x)\vee Q(y)))$; any $y$ at all will work.

Universal introduction has the condition that the name letter generalized upon, cannot occur in any assumption still in effect (equivalently, can't occur in any assumption in the assumption set). It doesn't matter if they appeared anywhere else, the rule only says that they can't appear in any assumption still in effect. At step 3 you have the name letters "v" and "w". So, can't generalize on v to get to line 4. This behaves just like the following:

1 | (Fa $\lor$ Gb) assumption

2 | $\forall$x(Fx $\lor$ Gb) 1 universal introduction (invalid!)

3 ((Fa $\lor$ Gb)$\implies$$\forall$x(Fx $\lor$ Gb)) 1-2 conditional introduction