You probably mean $\|f\| =\max{\{\|f\|_{\infty},\|f'\|_{\infty}\}}$ where $\|f\|_{\infty} = \sup_{t \in [0,1]} |f(t)|$.
Hint: Recall the basic fact that if $f_{n}$ is a sequence of $C^{1}$ functions such that $f_{n}$ converges uniformly (or only pointwise) to a continuous function $f$ and $f_{n}'$ converges uniformly to a continuous function $g$ then $f \in C^{1}$ and $f' = g$. It's probably a good idea to try to prove this on your own without consulting your books.
That $\|T\| \leq 1$ is just the definition of the operator norm. Consider also $f(x) = x$.
Let $ D_0 = \{ f \in L^2[0,1] | f \text{ is absolutely continous and } f'\in L^2[0,1] \} $ and $ T : D_0 \rightarrow L²[0,1], Tf=i\frac{d}{dt}f = i \cdot f'$.
Take a sequence $ \{ u_n \} $ which convergences to $ u $ in $ L^2 $- Norm and with $ { Tu_n } $ beeing convergent in $ L^2 $ with limit $ x $. We have to show that $ Tu = x $.
First we show that $ \{ u_n \} $ is uniform convergent. By Hölder's inequality
$$
\int_{0}^{t} {| i \cdot u_n'(s) - x(s)| ds} \le
\int_{0}^{1} {|i \cdot u_n'(s) - x(s)| ds} \le
\left( \int_{0}^{1} {ds} \right)^{\frac{1}{2}} \cdot \left( \int_{0}^{1} {|i \cdot u_n'(s) - x(s)|^2 ds} \right)^{\frac{1}{2}} = \| Tu_n - x \|_{L^2} $$
From this inequality we conclude that
$$ \| t \mapsto \int_{0}^{t} {i \cdot u_n(s)ds} - \int_0^{t}{x(s)ds} \|_{\infty} \le \| Tu_n - x \|_{L^2} $$
which gives uniform convergence of $ \{ t \mapsto \int_{0}^{t} i \cdot u_n(t) \} $ to $ t \mapsto \int_0^{t}{x(s)ds} $. Especially $ \{ t \mapsto \int_{0}^{t} i \cdot u_n(s)ds \} $ is a uniform Cauchy Sequence. Further we have
$$
| i \cdot u_n(0) - i \cdot u_m(0) |
= \left( \int_{0}^{1} | i \cdot u_n(0) - i \cdot u_m(0) |^2 dt \right)^{\frac{1}{2}}
= \left( \int_{0}^{1} | i \cdot u_n(t) - i \cdot u_m(t) - \int_{0}^{t} {(i \cdot u_n'(s)- i \cdot u_m'(s))ds} |^2 dt\right)^{\frac{1}{2}}
\\ \le \left( \int_{0}^{1} | i \cdot u_n(t) - i \cdot u_m(t) |^2 dt \right)^{\frac{1}{2}} + \left( \int_{0}^{1} | \int_{0}^{t} { i \cdot u_n'(s)-i \cdot u_m'(s)ds} |^2 dt \right)^{\frac{1}{2}}
$$
which implies
$$
| i \cdot u_n(0) -i\cdot u_m(0) | \le
\| i \cdot u_n - i \cdot u_m \|_{L^2} + \| t \mapsto \int_{0}^{t} { i \cdot u_n'(s)-i \cdot u_m'(s)ds} \|_{L^2} \\ \le
\| u_n - u_m \|_{L^2} + \| t \mapsto \int_{0}^{t} { i \cdot u_n'(s)-i \cdot u_m'(s)ds} \|_{\infty} $$
Hence $ \{ i \cdot u_n(0) \} $ is a Cauchy Sequence. Because each $ u_n $ is absolutly continous we have
$$ i \cdot u_n (t) = i \cdot u_n(0) + \int_{0}^t{i \cdot u_n'(s)ds} $$
which shows that $ \{i \cdot u_n\} $ and $ \{u_n\} $ are uniform convergent.
Now we show $ Tu = x $. Define $ g(t) = \lim_{n \rightarrow \infty} \frac{1}{i} u_n(0) + \frac{1}{i} \int_{0}^{t}{x(s)ds} $. Then we have $ Tg = x $ and $ i \cdot g' = x $ a.e. in $ [0,1] $. But $ \{ u_n \} $ convergences to g uniformly. Hence $ u = g $ a.e. in $ [0,1] $ which gives $Tu = x $.
Note: A similar reasoning can be found in Werner's Funktionalanalysis.
Best Answer
Probably you don't want just the sup norm on smooth functions. Rather, topologize smooth functions by the family of seminorms given by sups of all derivatives. (This is a Frechet space, the projective limit topology on the Banach spaces $C^k[a,b]$ with sum of sups of derivatives up to order $k$.) Differentiation is continuous in that topology. (And differentiation is continuous from $C^k[a,b]$ to $C^{k-1}[a,b]$.)
A reason to understand that just the sup norm of values is not the right topology on smooth functions is that the space is not complete in that topology.