A problem about Linear Operator $X$ and $Y$ are Banach Spaces.$ T$ is a linear bounded operator from $X \to Y$. There exists a real number $c$ which is positive, such that for any $y$ belonging to $T(X)$, there exists a $x$ which satisfies that $Tx=y$,and $$||x||\leq c||y||.$$
The problem is to prove that $T(X)$ is a closed set in $Y$.
I know that to prove $T(X)$ is closed is just check the limit of a sequence in $T(X)$ is also in $T$, but I don't know how to use the condition $||x||\leq c||y||$.
 A: Denote $Z=T(X)$, consider bounded linear operator $S:X\to Z:x\mapsto T(x)$. From assumption $S$ is open. Consider factor operator $\widehat{S}:X/\ker S\to Z:x+\ker S\mapsto S(x)$. Since $S$ is an open map, then $\widehat{S}$ is an isomorphism. As $X$ is a Banach so does its factor $X/\ker T$, hence $T(X)=Z\cong$$X/\ker T$ is complete. Complete subspace of Banach space is closed.
A: 1.If $N(T)=0$,the assumption says that $||x||\leq c||Tx||$ for all $x \in X$. Let $\{Tx_{n}\}$ be a sequence in $R(T)$ such that $Tx_{n}\rightarrow y$,we have to show that $y\in R(T)$.In fact ,since $$||x_{n}-x_{m}||\leq c||Tx_{n}-Tx_{m}||,$$ 
we see that $x_{n}$ is a Cauchy sequence in X,so there is a $x \in X$ such that $x_{n}\rightarrow x$.It follows that $$Tx_{n}\rightarrow Tx,$$since T is a linear bounded operator.Now we have $Tx=y$ which imply $y\in R(T)$.
2.If $N(T)\neq0$, define $\bar{T}:X/N(T)\mapsto Y$ as $\bar{T}\bar{x}=Tx$,it is easy to verify
that $X/N(T)$ is a Banach space with the norm $$||\bar{x}||=inf\{||z||:z \in \bar{x}\}$$ and $\bar{T}$ is a well-defined linear bounded operator.
Let $\bar{x} \in X/N(T)$,since $\bar{T}\bar{x}=Tx$,by the assumption,there exist a $z \in \bar{x}$ s.t. $||z||\leq c||Tx||$.Hence $$||\bar{x}||\leq||z||\leq c||Tx||=c||\bar{T}\bar{x}||,\forall \bar{x} \in X/N(T) $$
Now,by 1,$R(\bar{T})$ is closed $\Longrightarrow R(T)$ is closed. 
