Q:

closed range implies boundedness of operator

Let $X$ and $Y$ be a Banach spaces and let $T: X\rightarrow Y$ be linear, injective and bounded operator. Denote $R(T):= \left\{ Tx \ : \ x\in X\right\}$. Show that $T^{-1}: R(T)\rightarrow X $ is bounded iff $R(T)$ is closed.
I proved the implication from the right side to the left one. (If range is closed,then it is a Banach space, and using inverse mapping theorem we get the statement.)
How to prove this in the second direction?

A:

Suppose $T^{-1}$ is bounded. Then for some $M>0$ we have $\|T^{-1}y\|\leq M\|y\|$ for all $y\in R(T)$. This implies that $T$ is bounded below, which implies $R(T)$ is closed.  In case you're not familiar with this result, I'll include a proof:
Suppose $(y_n)$ is a Cauchy sequence in $R(T)$, and write $x_n=T^{-1}y_n$.  Given $\varepsilon>0$, there is some $N\in\mathbb N$ such that $\|y_n-y_m\|<\varepsilon/M$ for $n,m\geq N$. Then
$$\|x_n-x_m\|\leq M\|y_n-y_m\|<\varepsilon$$ 
for $n,m\geq N$, so $(x_n)$ is Cauchy, and thus convergent to some $x\in X$. Put $y=Tx$. Now there is some $N\in\mathbb N$ such that $\|x_n-x\|<\varepsilon/\|T\|$ for all $n\geq N$, and thus 
$$\|y_n-y\|\leq\|T\|\|x_n-x\|<\varepsilon$$
for $n\geq N$, and thus $y_n\to y$. Thus $R(T)$ is complete, and therefore closed.

