How to start proving $A \times B \times C \ne (A \times B) \times C$?

You didn't give a definition for $A\times B\times C$ but it's usually defined by $$A\times B\times C=\{(a,b,c) | a\in A, b\in B, c\in C\}$$ Now $(A\times B)\times C$ is, by definition $$(A\times B)\times C = \{((a,b),c) | (a,b)\in A\times B, c\in C\} = \{((a,b),c) | a\in A,b\in B, c\in C\}$$ This shows you the difference: The elements are of the form $((a,b),c) \ne (a,b,c)$, but they are, in fact, in a 1-1 correspondence, wich may be the point of confusion.

Addendum due to comment of Asaf: An ordered pair can be written as a set: $$(a,b) = \{a,\{a,b\}\}\\ (a,b,c) = \{a,\{a,b\},\{a,b,c\}\}\\ ((a,b),c) = \{(a,b),\{(a,b),c\}\} = \Big\{\{a,\{a,b\}\},\{\{a,\{a,b\}\},c\}\Big\}$$ Wich clearly shows why $((a,b),c) \ne (a,b,c)$.

Hint: What does a typical element of $(A \times B) \times C$ look like? A typical element of $A \times B \times C$?

A typical element of the former is $((a, b), c)$, which in particular is an ordered pair, and a typical element of the latter is $(a, b, c)$, which is an ordered triple. Still, as you point out, these sets are very similar, and in fact there is a canonical bijection between them, namely $$((a, b), c) \leftrightarrow (a, b, c).$$