The first thing to do is the write and understand the definitions of all the symbols in the equation.
Let us recall those:
- $\bigcup_\alpha A_\alpha=\{a\mid\exists\alpha.a\in A_\alpha\}$
- $\bigcap_\alpha A_\alpha=\{a\mid\forall\alpha.a\in A_\alpha\}$
- $A\setminus B=\{a\in A\mid a\notin B\}$
Now we can write a simple element chasing proof:
Let $x\in X\setminus\bigcap_\alpha Y_\alpha$. Then $x\in X$ and $x\notin\bigcap_\alpha Y_\alpha$, therefore for some $\alpha$, $x\notin Y_\alpha$, fix such $\alpha$. Therefore $x\in X\setminus Y_\alpha$, and therefore there exists $\alpha$ such that $x\in X\setminus Y_\alpha$, and by definition we have that $x\in\bigcup_\alpha (X\setminus Y_\alpha)$.
The other direction is as simple, take $x\in\bigcup_\alpha(X\setminus Y_\alpha)$, then for some $\alpha$ we have $x\in X\setminus Y_\alpha$. Therefore $x\in X$ and $x\notin Y_\alpha$, so by definition $x\in X$ and $x\notin\bigcap_\alpha Y_\alpha$, i.e. $x\in X\setminus\bigcap_\alpha Y_\alpha$.
The second identity has a similar proof. I like these proofs because they not hard and give a good exercise in definitions and elements chasing.