Question: I tried doing this by Eisenstein's criterion:

$2$ is prime in $\mathbb{Q}$ and I then proceeded to write that it divides $-2$, $-2$ and $0$ but doesn't divide $1$ and also $2^2=4$ doesn't divide $2$. I then noticed these last two are erroneous since we're in $\mathbb{Q}$ and we have that $(1/2)(2)=1$ and thus $2$ does divide $1$; also $(4)(1/2)=2$ so $4$ divides $2$.
How could I have done it instead?

Answer: 
You solve it by noting that the polynomial is monic, which means that a factorisation over the rationals gives a factorisation over the integers (or contrapositively: of the is no non-trivial factorisation over the integers, then there are none over the rationals). And over the integers you can use Eisenstein without any issues.
Suppose that $p/q\in\mathbb Q$ is a root of this polynomial, with $p,q$ coprime integers. Then you have that
$$\frac{p^3}{q^3}-2\frac{p}{q}-2=0$$
or, by doing a bit of algebra,
$$p^3=2q^2 (p+q)$$
from which you can derive a contradiction.