I know how to prove that $x^n$ is not cauchy in $(C[0,1],\|\cdot\|_\infty)$, but my question is that since $(C[0,1],\|\cdot\|_\infty)$ is complete, and the pointwise limit on $x^n$ is:
$x^n\to f(x) = \left\{ \begin{array}{ll} 1 & \mbox{if $x =1$};\\ 0 & \mbox{if $0\le x \lt 1 $}.\end{array} \right. $
And this limit is not continuous, and thus is not in $C[0,1]$.
So if the sequence were cauchy then the limit would exist in $C[0,1]$ (by completeness), so isn't that an instant contradiction? Meaning no formal proof is necessary?
Yes you are absolutely correct. But proving that $C[0,1]$ is complete with respect to $L_\infty$ norm is at least a little bit non-trivial.