Proving $x^3$ is Big-O of $x^2$ as $x \rightarrow 0$

EDIT: I have attempted the question again and posted an answer in a separate post if you scroll blow. Can you please check my NEW working?

The definition I have is that $f(x)$ is Big-O of $g(x)$ as $x \rightarrow x_0$ if there exists $K, \delta >0$ such that $|f(x)| \leq K |g(x)|$ when $0<|x-x_0|< \delta$.

In this case $f(x)=x^3$, $g(x)=x^2$ and $x_0=0$.

I need to show that there exist $K, \delta >0$ such that if $0 < |x| < \delta$, then $|x^3| \leq K |x^2|$.

I know that if $\delta < 1$, then $|x^3| < |x^2|$, which gives $0 < |x| < \delta$ implies that $|x^3| \leq |x^2| = 1 \cdot |x^2|$ where $K=1$

Does that mean I have proved this as there exists a $K=1 > 0$ where this is true and there exists a $\delta > 0$ where this is true, namely $0 < \delta < 1$?