Assume that $0<a<1$ and let $X$ be a closed subspace of $C[0,1]$ such that each $f\in X$ is continuously differentiable on $[0,1)$.
I am trying to show that the operator (it might not be) $D: X \rightarrow C[0,a] $ given by $D(f) = f'|_{[0,a]}$ is a closed graph.
My idea was to take a sequence $(f_n)_{n=1}^\infty \rightarrow 0$ in $X$ and $(f_n' \rightarrow y)$ in $C[0,a]$ and show that $y=0$. But at the moment I am stuck.
On
Your space $X$ is contained in the Fréchet space $Y=\{f\in C[0,1]: f|_{[0,1)} \in C^1[0,1)\}$ which is endowed with the semi-norms $\|f\|_n=\sup\{|f(x)|: x\in[0,1]\}+\sup\{|f'(x)|: x\in[0,1-1/n]\}$. Clearly, $Y$ is continuously embedded in the Banach space $Z=C[0,1]$ and since $X$ is closed there it is also closed in $Y$. The closed graph theorem implies that the topologies of $Y$ and $Z$ coincide on $X$. This implies the continuity of $X\to C[0,a]$, $f\mapsto f'|_{[0,a]}$.
What you say your idea was is not what you need to do to show the graph is closed! You need to show this:
If $f_n\to f$ in $X$ and $f_n'\to g$ in $C[0,a]$ then $f'=g$.
Which is precisely a result from advanced calculus: If $f_n\to f$ uniformly and $f_n'\to g$ uniformly then $f'=g$. Seems to me the simplest proof is to note that $$f_n(x)-f_n(0)=\int_0^x f_n'(t)\,dt,$$so $f_n'\to g$ uniformly implies $$f(x)-f(0)=\int_0^x g(t)\,dt,$$hence $f'=g$.