How does 'AND' distribute over 'OR' (Set Theory)?

The reason "similarly" was written is because all the $\implies$ could be replaced with $\iff$, thus allowing a path from $x\in(A\cup B)\cap(A\cup C)$ to $x\in A\cup(B\cap C)$ by reading backwards. Thus, you actually need no effort to prove the reverse case; it has already been printed out for you.

The set operations $\cap$ and $\cup$ correspond exactly with the $\land$ and $\lor$ of Boolean logic.

Remember that Distribution goes two ways. That is, you can go from $X \land (Y \lor Z)$ to $(X \land Y) \lor (X \land Z)$, but you can go from $(X \land Y) \lor (X \land Z)$ back to $X \land (Y \lor Z)$.

Now, going the other way doesnt feel like 'Distribution' (it feels more like a 'Reverse Distribution' or 'Çollecting common Terms'), which is exactly why so many beginning students of logic miss it, and instead end up doing the exact same thing you do: going from $(X \land Y) \lor (X \land Z)$ to $((X \land Y) \lor X) \land ((X \land Y) \lor Z)$ ... But that, as you saw, doesn't really go anywhere. This is a very common 'mistake'!

So, the key is to do Distribution 'the reverse way' after line 2:

$$ \begin{aligned} &y\in(A\cup B) \land y\in(A\cup C)\\ \implies &(y\in A \lor y\in B) \land (y\in A \lor y\in C)\\ \implies &y\in A \lor (y\in B \land y\in C)\\ \implies &y\in A \lor (y\in B \cap C)\\ \implies &y\in A \cup (B \cap C) \end{aligned} $$

And this, as @ParclyTaxel notes, just with all steps reversed from the first proof.

Indeed, it is always true that $(\alpha\lor\beta)\land(\alpha\lor\gamma)\equiv\alpha\lor(\beta\land\gamma)$. You can see this by verifying it semantically using truth tables (so there is also a derivation for it, but I can't seem to remember it):

\begin{array}{|c|c|c|c|} \hline \alpha&\beta &\gamma & (\alpha\lor\beta)\land(\alpha\lor\gamma) &\alpha\lor(\beta\land\gamma) \\ \hline 0& 0& 0& 0 &0\\ \hline 0& 0& 1& 0 &0\\ \hline 0& 1& 0& 0 &0\\ \hline 0& 1& 1& 1 &1\\ \hline 1& 0& 0& 1 &1\\ \hline 1& 0& 1& 1 &1\\ \hline 1& 1& 0& 1 &1\\ \hline 1& 1& 1& 1 &1\\ \hline \end{array}

We can thus write the expression

$$(y\in A\lor y\in B)\land (y\in A\lor y\in C)$$

as

$$y\in A\lor (y\in B\land y\in C)$$

which gives your claim and you don't have to expand any further.