Let $X$, and $Y$ be topological vector spaces and let $D$ be a dense vector subspace of $X$.
An operator $T:D\to Y$ is called closed iff the graph of $T$, $\{(x,T(x))\in X\times Y|\,x\in D\}\subseteq X\times Y$ is a closed subset (in the product topology).
A map $T:D\to Y$ between two topological spaces is closed iff $T(C)\in \operatorname{Closed}(Y)$ for all $C\in \operatorname{Closed}(D)$ where $D$ is taken with the subspace topology.
Question: Are these two notions equivalent? If no, why is there a discrepancy in the terminology?
Attempt at answer: No. Take $X$ and $Y$ to be Banach spaces, let $(f_n,T(f_n))_{n\in\mathbb{N}}$ be a sequence in the graph of $T$ which converges to some point $(f,g)\in X\times Y$. In order for the graph of $T$ to be closed, we need $f\in D$ and $T(f)=g$. Then there is no way to apply the fact that $T$ is a closed map in order to conclude $f\in D$, right?
As you found out, the two notions are not equivalent. It happens often that two different areas of mathematics use the same word to mean two different things. Think of all the meanings of "normal", "complete", or "regular"...
The map $T:\ell_2\to\ell_2$ defined by $T(\{x_n\}) = \{x_n/n\}$ has a closed graph, but does not map closed sets to closed sets: indeed, the range of $T$ is a proper dense subspace of $\ell_2$.
On the other hand, the map $T(x)=0$, $x\in c_{00} \subset \ell_2$, sends closed sets to closed sets without having a closed graph. [Example suggested by Daniel Fischer].