I know if $X$ were to incomplete space such as $c_{00}$
the hypothesis does not hold but what about if $X$ were to be a Banach space?
My thought is $R(T)$ would be a Banach space too because it is a closed subset of a complete metric space. Therefore for $x_n\to x$ and $Tx_n\to y$ since the range is complete $y=Tx$ (but I think I need injectivity of $T$, as well.)
(Note that you are asking a different question than the one you linked to, here you are asking if $T$ maps closed sets to closed sets, does $T$ have a closed graph? In the other question you are asking if $T$ has a closed graph, does it map closed sets to closed sets?)
First note that if $T:X\to Y$ is a linear operator between normed spaces that sends closed sets to closed sets, then the kernel of $T$ is either $0$ or all of $X$ (ie $T=0$). For suppose $x\notin\ker(T)$ and $y\in\ker(T)$ with $\|y\|>\|x\|$. Then $A=\{\frac1n x + 100 n^2 y\mid n\in\Bbb N\}$ is discrete in $X$, hence closed. But its image is $\{\frac1n T(x)\mid n\in\Bbb N \}$ which converges to $0$, but $0$ does not lie in $T(A)$, contradiction.
So your map must either be the zero map or be injective. If its the zero map it is continuous, if it is injective then $T(X)$ is a closed subspace of $Y$. Then $T: X\to T(X)$ is a bijective map sending closed sets to closed sets, so it is an open map. If $X,Y$ is Banach then $T(X)$ and $X$ are Banach, so $T: X\to T(X)$ is an open map between Banach spaces, hence an isomoprhism and as such continuous. Continuous implies that the graph is closed.
