Let $X, Y$ be normed spaces and $D \subseteq X$ a subspace. Let $T: X \supset D \to Y$ and $Tx=0$ for any $x$. Is $T$ closed?
I do not believe that it is closed, consider $c_{c}$ the space of sequences with finite support. $c_{c}$ is not closed in $\ell^{1}$ for example (since $c_{c}$ is dense in $\ell^{1}$) and thus we can find a sequence where $x \in \ell^{1}$ and $(x_{n})_{n} \subseteq c_{c}$ so that $x_{n} \to x$, while $Tx_{n}=0$ for all $n \in \mathbb N$. But by defintion $x \notin \operatorname{dom}(T)$ and hence we are done?
I do not know whether I have truly understood this exercise, it seems to simple
A closed map is a map which maps closed sets into closed sets. If $C\subset D$, then $T(C)=\emptyset$ (if $C=\emptyset$) or $T(C)=\{0\}$ (otherwise). In each case you get a closed set (even without assuming that $C$ is closed) and therefore, yes, $T$ is a closed map.
Besides, the graph of $T$ is a closed subset of $D\times Y$.
Edit: On the other hand, if you had in mind that $T$ may be a closed unbounded operator, then you are right. If $D$ is not a closed subset of $X$, then the graph of $T$ is not a closed subset of $X\times Y$. Therefore, in that case (as in your example), $T$ will not be closed.