For (1): Another condition for $||\cdot||$ to be a norm is being positive definite, i.e. $||f|| = 0 \iff f = 0$. Once you check this as well, $||\cdot||$ will be a norm by definition.
For (3): There is no need to check if $C[0,1]$ is closed, since in a topological space $X$, the set $X$ will always be closed. Only completeness needs to be shown in order to be a Banach space. But in this case, the normed space is not complete! Look at the following counterexample:
$$f_n(x) = \begin{align} \begin{cases} nx \hspace{1cm} \text{if} \hspace{0.2cm} 0 \leq x \leq \frac{1}{n} \\ 1 \hspace{1cm} \text{else} \end{cases} \end{align}.$$
It's easy to see this converges pointwise to $f(0) = 0$ and $f(x) = 1$ for $x > 0$. Now it's not hard to show that if the sequence would converge, then it would converge to this $f$, but $f$ is not continuous, giving a contradiction.
The problem with using the Arzelà-Ascoli theorem is that not all conditions of this theorem are satisfied. For example, a Caushy sequence $f_n$ in $C[0,1]$ need not be uniformly bounded (if you want a counterexample, take inspiration from the previous $f_n$), nor equicontinuous. Also if a subsequence of a Caushy sequence converges, that does not imply that the whole sequence converges (this does hold however if we take the norm $||f||_{\infty} = \sup_{x \in [0,1]} |f(x)|$).
Best Answer
This is false. Let $C$ be a 'fat' Cantor set of positive measure. Since $C$ is closed we can find continuous functions $f_n$ with values in $[0,1]$ such that $f_n \to I_C$ pointwise. By Bounded Convergence Theorem $f_n \to I_C$ in $L^{2}[0,1]$. But $I_C$ is not piecewise continuous.