This is a proof that i think is wrong that i found online.
Assume that the image is closed, then the map $l_1 \to T(l_1)$ Is a linear bijection. Thus by open mapping theorem $T(l_1)$ is both open and closed in $l_2$ and so the image must be the entire $l_2$ but there is no bijection between $l_1$ and $l_2$. I think the mistake is to claim $T(l_1)$ is both open and closed in $l_2$. I think it is open and closed in the subspace topology on $T(l_1)$ but that is all. I think the correct solution is to note that $T(l_1)$ is a Hilbert space and so it cannot be isomorphic to $l_1$. Are my notes correct?