Prove $(\cup a)\cup (\cup b)=\cup(a\cup b)$

Question

I'm learning ZFC set theory and I'm stuck in this question: $$ (\cup a)\cup(\cup b)=\cup(a\cup b). $$ Here $a,b$ are given sets.

MY PROGRESS

Here are my steps of proof:

\begin{equation*} \begin{split} \forall x(x\in (\cup a)\cup (\cup b) &\leftrightarrow x\in \cup a\vee x\in \cup b\\ &\leftrightarrow \exists u(u\in a\wedge x\in u)\vee \exists v(v\in b\wedge x\in v)) \end{split} \end{equation*} and \begin{equation*} \begin{split} \forall x(x\in \cup (a\cup b) &\leftrightarrow \exists w((w\in a\vee w\in b)\wedge x\in w)\\ &\leftrightarrow \exists w((w\in a\wedge x\in w)\vee(w\in b\wedge x\in w))). \end{split} \end{equation*} If $$ (\exists up(u)\vee \exists vq(v))\leftrightarrow \exists w(p(w)\vee q(w))\tag{1} $$ is true for all formulae $p,q$, then everything is done, but I can only understand (1) intuitively without proving it formally.

By the way, if (1) becomes $$ (\exists up(u)\to \exists vq(v))\leftrightarrow \exists w(p(w)\to q(w)), $$ then it seems false --- also remain unproved.

I AM ALLOWED TO USE

1. All about propositional, predicate logic and proof system. The allowed logical axioms are tautologies, two quantifier axioms and axioms of equality. The allowed laws of inference are modus ponens and generalization.

2. First five ZF axioms (without the axioms of infinity, replacement, regularity, and choice)

Answer 1

$$x\in \bigcup a \cup \bigcup b \iff (\exists u\in a, x\in u) \vee(\exists v\in b, x\in v)$$

We note that $u\in a\subseteq a\cup b$ and $u\in b\subseteq a\cup b$, so $u,v\in a\cup b$. Thus, we have, in either case that

$$\iff \exists w\in a\cup b, x\in w$$

$$\iff x\in \bigcup(a\cup b)$$

Now, to answer your question about the propositional form $(\exists u\in a, p(u)\vee \exists v\in b, q(v))\iff \exists w\in a\cup b(p(w)\vee q(w))$: We are dealing with a very special pair of $p,q$. Namely, realize that $p \equiv q$!! In fact, both denote membership... $p(u)\equiv x\in u$ and $q(v) \equiv x\in v$. So the actual propositional form is just

$(\exists u\in a, p(u)\vee \exists v\in b, p(v))\iff \exists w\in a\cup b, p(w))$

which is a tautology.