I'm reading the first section of chapter 4 of Mei Chi's book, and I do not follow a part of a proof. Let $T:H_1\rightarrow H_2$ be a linear, closed, densely defined oeprator. Then the range of $T$ is closed if and only if there is a constant $C$ such that $||f||_1\leq C ||Tf||_2$ for all $f\in Dom(T)\cap \overline{R(T^*)}$
I understand the forward direction. However, I do not understand how to prove the backward direction. The author says it is obvious. How can I see that it is obvious that the range of $T$ is closed given the inequality above?
Let $\{Tf_j\}$ be a Cauchy sequence in $R(T)$. Then $$ \|f_j-f_k\|\leq C\,\|T(f_j-f_k)\|=C\,\|Tf_j-Tf_k\|. $$ So $\{f_j\}$ is Cauchy. As $H_1$ is complete, there exists $f=\lim f_j$. Write $y=\lim Tf_j$ (this exists because the sequence is Cauchy and the space complete). For any $h\in D(T^*)$, $$ \langle Tf,h\rangle=\langle f,T^*h\rangle=\lim_j\langle f_j,T^*h\rangle=\lim_j\langle Tf_j,h\rangle=\langle y,h\rangle. $$ Then $$\tag1\langle y-Tf,h\rangle=0$$ for all $h\in D(T^*)$. Because $T$ is densely defined and closed, we have $R(T)\subset D(T^*)$. Thus we may take $h=y-Tf$ in $(1)$ to obtain $y-Tf=0$. So $y=Tf\in R(T)$ and so $R(T)$ is closed.