Question: I need to show that the cross product is not associative without using components.
I understand how to do it with components, which leads to an immediate counterexample, but other than that I am not sure how to do it.

Answer: 
Consider two non-zero perpendicular vectors $\def\v#1{{\bf#1}}\v a$ and $\v b$. We have
$$(\v a \times\v a)\times\v b=\v0\times\v b=\v0\ .$$
However $\v a\times\v b$ is perpendicular to $\v a$, and is not the zero vector, so
$$\v a\times(\v a\times \v b)\ne\v 0\ .$$
Therefore
$$(\v a \times\v a)\times\v b\ne\v a\times(\v a\times \v b)\ .$$
The cross product does satisfy the Jacobi identity
([vector algebra relations](https://en.wikipedia.org/wiki/Vector_algebra_relations#Addition_and_multiplication_of_vectors))

$$ a \times (b \times c) +
 b \times (c \times a) +
 c \times (a \times b)
 = 0.$$

Move the 3rd term to the right
and use anti-commutativity ($\,u \times v = -v \times u\,$) to get

$$ a \times (b \times c) +
 b \times (c \times a) =
 (a \times b) \times c. $$

However the term
$\, b \times (c \times a) \,$
is not always zero which prevents
associativity.