Show that $C^1[0,1]$ is not a banach space using the closed graph theory with the maximum norm. First, look at the derivative operator: $D:C^1[0,1]\to C[0,1]$, $D(f)=f'$.
We can check that $D$ is linear and not bounded (by taking an example such as a polynomial $x^{n+1}$). Thus $D$ is not continuos. I'm not sure, if it is possible to show that $D$ has a closed graph (a linear map $T:X\to Y$ has a closed graph if $x_n\subset X$ such that $x_n\to x$ and $T_{x_n}\to y$ then $Tx=y$).So if by contradiction, we assume that $C^1[0,1]$ is banach with the sup norm, then get by the closed graph theory that $D$ is continuos, which is not true according to what we've said.
$D$ has a closed graph: Let $u_n \to u$ and $D(u_n) = u_n' \to v$ uniformly. We want to show that $v=D(u)=u'$. To this end, notice that for all $t \in [0,1], \ \int_0^t u_n'(s) ds \to \int_0^t v(s) ds$ and thus $ u_n(t) - u_n(0) \to \int_0^t v(s) ds$. In conclusion, $u(t) =u(0) + \int_0^t v(s) ds$ and so $ u \in C^1[0,1]$ and $u'=v$.