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.
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
$L$ is closed; that is, $L$ has closed graph: If $u_n$ converges uniformly to $u$ and $Lu_n=u_n'$ converges uniformly to $f$, it follows that $Lu=u'=f$. See this, for example.
The mistake in the argument is that you can't appeal to the Closed Graph Theorem, as the (needed) hypotheses that the domain space be complete is not satisfied by your $X$. (For instance, there is a sequence of polynomials converging uniformly to the function $x\mapsto |x-1/2|$ on $[0,1]$ by the Weierstrass Approximation Theorem.)