Prove, by the definition of continuity, that $f(x)=ax^n$ is continuous I would really appreciate an extra pair of eyes or two to see if I missed any details here. 
Proof. We will break the proof into a quadotomy. 
i) Suppose $n=1$, then $\forall \ \varepsilon>0$, and $\forall \ x_0\in\mathbb{R}$, pick $\delta = \varepsilon/|a|$ so that $|x-x_0|<\delta$ implies
$$|f(x)-f(x_0)|=|ax-ax_0|=|a||x-x_0|<|a|\delta=|a|\varepsilon/|a|=\varepsilon.$$
So, $f$ is continuous for $n=1$, and we can assume $n\geq 2$. 
ii) Suppose $n\geq 2$ and $x_0=0$, then $\forall \ \varepsilon>0$, pick $\delta=\sqrt[n]{\varepsilon/|a|}$ so that $|x-0|<\delta$ implies
$$|f(x)-f(0)|=|ax^n-a0^n|=|a|\cdot|x|^n<|a|\delta^n=|a|(\sqrt[n]{\varepsilon/|a|})^n=\varepsilon.$$
So, $f$ is continuous for $n\geq 2$ and $x_0=0$ we can assume $|x_0|>0$.
iii) Suppose $n\geq 2$ and $x_0>0$, then $\forall \ \varepsilon>0$, pick $\delta=\min\{\varepsilon/(K\cdot|a|),x_0/2\}$, where $K=x_0^{n-1} \frac{1-(3/2)^n}{1-3/2}$, so that $|x-x_0|<\delta$ implies
$$|f(x)-f(x_0)|=|ax^n-ax_0^n|=|a|\cdot|x-x_0|\cdot |x^{n-1}+x^{n-2}x_0+\cdots+xx_0^{n-2}+x_0^{n-1}|$$
since $|x-x_0|<x_0/2$, we know $x_0/2<x<3x_0/2$, then
$$<|a|\cdot|x-x_0|\cdot[(3x_0/2)^{n-1}+x_0(3x_0/2)^{n-2}+\cdots+x_0^{n-2}(3x_0/2)+x_0^{n-1}]$$
$$=|a|\cdot|x-x_0|\cdot x_0^{n-1}[(3/2)^{n-1}+(3/2)^{n-2}+\cdots+3/2+0]$$
$$=|a|\cdot|x-x_0|\cdot x_0^{n-1}\sum\limits_{j=0}^{n-1} \left(\frac{3}{2}\right)^j=|a|\cdot|x-x_0|\cdot x_0^{n-1}\frac{1-(3/2)^n}{1-3/2}=K\cdot|a|\cdot|x-x_0|$$
$$<K\cdot|a|\cdot\delta\leq K\cdot |a|\cdot \frac{\varepsilon}{K\cdot |a|}=\varepsilon$$
So, $f$ is continuous for $n\geq 2$ and $x_0$.
iv) It can be shown using a similar method to iii) that $f$ is continuous for $n\geq 2$ and $x_0<0$.
Thus, $f(x)=ax^n$ is continuous for all $x\in\mathbb{R}$. $\square$
 A: Looks very good from what I can see. Very rigorous and exact. There is absolutely nothing wrong with it.

However, let me give another example which is just as strict, but perhaps (in my completely subjective opinion), slightly easier to follow.
For case iii:
Let $M= \max_{x\in(x_0-1, x_0+1)}\left\{|a||x^{n-1} + x^{n-2}x_0 + \cdots + x_0^{n-1}|\right\}$.
Pick $$\delta = \min\left\{1, \frac{\epsilon}{M}\right\}.$$
Now, assuming that $|x-x_0|<\delta<1$, we know that $x\in(x_0-1, x_0+1)$. Therefore, 
$$|a||x^{n-1} + x^{n-2}x_0 + \cdots + x_0^{n-1}| < \max_{x\in(x_0-1, x_0+1)}\left\{|a||x^{n-1} + x^{n-2}x_0 + \cdots + x_0^{n-1}|\right\} = M.$$
Now, we can simply see that 
$$|f(x)-f(x_0)| = |a||x-x_0||x^{n-1} + x^{n-2}x_0 + \cdots + x_0^{n-1}|< M\cdot\delta <M\cdot\frac\epsilon M = \epsilon.$$

To me, the proof above is slightly easier to follow, as it in effect first limits itself to looking at $f$ on $(x_0-1, x_0+1)$ - as continuity really only cares for values close to $x_0$.
However, like I said, this proof is not more correct than yours, it's purely subjective opinion.
