$A\cup B = (A\setminus B) \cup (B\setminus A) \cup (A\cap B)$. These three sets are disjoint, so
$$
|A\cup B| = |A\setminus B| + |B\setminus A| + |(A\cap B)|
$$
But $A\setminus B = A\setminus(A\cap B)$, so $|A\setminus B|=|A|-|A\cap B|$. A similar equality holds for $|B \setminus A|$. Substitution of these into the displayed equation above yields your result.
Of course, one might need to formally show that $|A\setminus (A\cap B)| = |A|-|A\cap B|$. I can't decide if this is any less obvious than the original proposition...
Venn diagrams are not a formal proof, nor a substitute for it, just an illustrative tool that can be useful as a guiding tool for your narrative/proof.
If writing a formal proof for this law, you will need to show
$$A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C) \;\;\; \text{and} \;\;\; (A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)$$
and then use the fact that if $X \subseteq Y$ and $Y \subseteq X$, then $X = Y$.
If you don't need formality, then in the appropriate context it can be used, I suppose. And, up to your ability to produce said diagrams, you could use a Venn diagram of $n$ circles, depending on what you're proving, but it gets messy quick so I wouldn't recommend it for more than $3$ sets.
In short, it depends on the level of formality that is expected of you. There's no denying that Venn diagrams in contexts like these are super, super helpful in illustrating concepts, and can be taken as a sort of heuristic proof, but they are not a substitute for formal proofs.
I say this in light of the assumption that you are probably encountering this in a class of some sort like a number of questions here. Classes in set theory, generally, will expect formality, not Venn diagrams, for example. In research, publications, journals, etc., things are much, much murkier depending on the context.
Best Answer
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:
Or you can try to do both inclusions at the same time using equivalences. Then you would have something like
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.