Consider the subspace $$D=\left\{x\in \ell^2 \ \big|\ \sum_{n\in\mathbb N} n^2 |x_n|^2<\infty\right\}$$ of $\ell^2$, and let $T:D\to\ell^2$ be defined by $T(\{x_n\})=\{n x_n\}$.
I need hints to prove that the graph of $T$ is closed.
Progress
To show $T$ has a closed graph, we proceed as follow: Assume the sequence $\{x^{(m)} \}$ in $D$ such that $x^{(m)}$ converges to $x$ as $m\to\infty$ for some $x\in\ell^2$ and assume $Tx^{(m)}$ converges to $y$ as $m\to\infty$. Our aim is to show $x\in\ell^2$ and $y=Tx$.
You don't need to show $x\in \ell^2$; that is a part of the set-up.
The equality $y=Tx$ can be checked component-wise: you need to check that $y_n=(Tx)_n$ for every $n$. That is, $y_n=nx_n$. Here are the two steps leading to the goal:
$$x_n =\lim_{m\to\infty} x^{(m)}_n\tag{1}$$ $$y_n =\lim_{m\to\infty} (Tx^{(m)})_n\tag{2}$$