Proving "by element-wise" (I do hope you are translating; that's awful abuse of language) means showing that each side is a subset of the other. In other words, the argument looks something like:
Suppose $x\in A\cup(B\cap C)$; then .... stuff ....
, so $x\in (A\cup B)\cap (A\cup B)$; therefore, $A\cup(B\cap C) \subseteq (A\cup B)\cap (A\cup C)$.
Now suppose that $y\in (A\cup B)\cap (A\cup C)$. Then ... stuff...
, so $y\in A\cup (B\cap C)$. Therefore, $(A\cup B)\cap (A\cup C) \subseteq A\cup(B\cap C)$.
Since each is contained in the other, $A\cup(B\cap C)=(A\cup B)\cap(A\cup C)$. QED.
Or you can try to do both inclusions at the same time using equivalences. Then you would have something like
$x\in A\cup (B\cap C) \Longleftrightarrow$ stuff
$\Longleftrightarrow$ more stuff
$\Longleftrightarrow$ even more stuff
$\Longleftrightarrow x\in (A\cup B)\cap (A\cup C)$.
In other words, "element-wise" means "chasing an element": pick an element in one side, show it has to be in the other and vice versa. After all, equality of sets is defined in terms of them containing the same elements, not in terms of Venn diagrams.