Proof that $f(x)=x^{1/n}$ is continuous. Here's what I've done:
According to the definition, a function is continuous at $c$ if, for any $\epsilon>0$, there exists a $\delta>0$ so that, if $|x-c| < \delta$, then $|f(x)-f(c)| < \epsilon$.
$$\begin{split}
|f(x)-f(c)| < \epsilon
  & \Leftrightarrow \left|x^{1/n}-c^{1/n}\right| < \epsilon \\
  & \Leftrightarrow c^{1/n}-\epsilon < x^{1/n} < c^{1/n}+\epsilon \\
  & \Leftrightarrow \left(c^{1/n}-\epsilon\right)^n < x < \left(c^{1/n}+\epsilon\right)^n,
\end{split}
$$
(which we can call $a < x < b$).
Thus, if we make $\delta = \min\{c-a,b-c\}$, then
$$|x-c| < \delta \Rightarrow |f(x)-f(c)| < \epsilon$$
Which proves its continuity. Have I done anything wrong?

I think you want to take $\delta = \min\{(c^{1/n}+\epsilon)^n-c, c-(c^{1/n}-\epsilon)^n\}$ for $c\ne0$. Otherwise just take $\delta=\epsilon^n$.
EDIT: Looks like you made the correction as I was posting.
