Compactness of $C([0,1])$ I have to verify if the $C([0,1])$, space of all continuous functions defined on interval $[0,1]$ with supremum metric is compact.

As I know, we have to check if every sequence of functions $f_{n}(x)$ has subsequence that $f_{n_{k}}(x)$ is convergent.
In this metric of course conervgence implies uniform convergence, so there won't be a problem of showing continuity of the limit function.
But I really don't know if we can make subsequence convergent.
The sequence $f_n(x)=x^n$ does not have a convergent subsequence since $f_n$ converges pointwise towards $f(x)=0$ if $x\neq 1$ and $f(1)=1$ which is not continuous. Hence, the space is not compact.
A Banach space is finite-dimensional if (and only if) it is locally compact. It's easy to see that $C[0,1]$ is a Banach space with respect to the supreme norm. And, trivially, the property of compactness implies local compactness.
To sum it up, there is not infinite-dimensional Banach space (such as $C[0,1]$ ) that is also compact.