How do I get $\|x\|\le C\|y\|$ in this case? I feel that the title is a bit uninformative, please feel free to edit it.
This is a problem related to the Open Mapping Theorem. Let $T:X\to Y$ be a bounded linear operator from a Banach space X to a Banach space Y. Suppose that there exist a constant $C>0$ such that for any $y\in D\subset Y$, $D$ is dense in $Y$, these conditions are satisfied 

  
*
  
*$\exists x\in X$ such that $Tx=y$.
  
*$\|x\|\le C\|y\|$.
  

I am trying to show that the result holds for any $y\in Y$, the closure of $D$.
Let $y\in Y$ be an arbitrary element, then we can write $y$ as 
$$
y=\sum_{n=1}^{\infty}y_n
$$
where $y_n\in D$ for each $n\in \Bbb N$. We can chose $(y_n)$ so that 
$$
\sum_{n=1}^{\infty}\|y_n\|<\infty
$$
since $Y$ is Banach. For each $n$, we let $x_n\in X$ be an element such that $Tx_n=y_n$ and $\|x_n\|\le C\|y_n\|$. Then 
$$
\sum_{n=1}^{\infty}\|x_n\|\le \sum_{n=1}^{\infty}C\|y_n\|<\infty
$$
by our assumption, thus $x=\sum_{n=1}^{\infty}x_n\in X$ since $X$ is Banach.
It's not hard to see that 
$$
Tx=T(\sum_{n=1}^{\infty}x_n)=\sum_{n=1}^{\infty}y_n=y
$$
but this is where I got stuck. I can't show that $\|x\|\le C\|y\|$. Can  anyone please suggest me an idea on how to proceed? An alternative proof would be fine too if you can explain how my method is doomed to fail.
 A: Solution 2 (without assuming $D$ to be linear) 
This is continuation of your solution: you got to the point where $\forall y\in Y\,\exists x\in X: Tx=y$ and therefore $T$ is an open mapping. This means that $T$ maps the open unit ball $B_X(0,1)\subset X$ in an open set in $Y$ which contains $0_Y\Rightarrow\exists r>0:\,\overline{B_Y(0,r)}\subset T(B_X(0,1))$. From here we see that $\forall y\in Y:\|y\|\leq r\,\exists x\in X: \|x\|\leq 1$ and $ Tx=y$. Now for arbitrary $y\in Y$ take the element $\frac{r}{\|y\|}y$ which has norm $r\Rightarrow \exists x\in B_X(0,1): Tx=\frac{r}{\|y\|}y\Leftrightarrow T(\frac{\|y\|}{r}x)=y$. Set $u:=\frac{\|y\|}{r}x\Rightarrow \|u\|=\|\frac{\|y\|}{r}x\|\leq \frac{1}{r}\|y\|$ and so the constant is $C=\frac{1}{r}$.
A: Solution 3 (shortest and continuation of your solution)
For each $y\in Y$ you can construct the series $y=\sum\limits_{n=1}^{\infty}{y_n}$ such that $\sum\limits_{n=1}^{\infty}{\|y_n\|}\leq 2\|y\|$ (just choose the $\epsilon_n$ from your comment as $\epsilon_n=\frac{\|y\|}{2^{n+1}}$).
$$\|x\|=\lim\limits_{N\to\infty}{\|\sum\limits_{n=1}^{N}{x_n}\|}\leq\lim\limits_{N\to\infty}\sum\limits_{n=1}^{N}{\|x_n\|}\leq \lim\limits_{N\to\infty}{C\sum\limits_{n=1}^{N}{\|y_n\|}}\leq 2C\|y\|$$ and so the constant for the whole $Y$ is $2C$.
A: We assume $D$ to be a subspace.
If $T$ is injective, then
 let $y_n\to y\in Y$ with $\{y_n\}\subset D$. Then $\exists \{x_n\}\subset X: Tx_n=y_n$ and $\|x_n\|\leq C\|y_n\|\Rightarrow \|x_n-x_m\|\leq C\|y_n-y_m\|\,(\text{because $T$ is injective})\,\Rightarrow$ the sequence $\{x_n\}$ is Cauchy in $X$ and therefore $\exists x\in X: x_n\to x$. But then $Tx_n\to Tx$ and we also have $y_n\to y\Rightarrow Tx=y$. Also $\|y_n\|\ge \frac{\|x_n\|}{C},\,\forall n\in\mathbb N\Rightarrow \|y\|\ge \frac{\|x\|}{C}$.
If $T$ is not injective, then you can consider the operator $\overline T: X/\text{Ker } T\to Y$, which has exactly the same image in $Y$ as $T$. In more details, you factorize $X$ with respect to the kernel of $T$ and give a norm to the space $X/\text{Ker } T$ as follows: $\|[x]\|=\|x+Ker T\|=\inf\limits_{z\in Ker T}{\|x-z\|}$. With this norm, the space $X/ Ker T$ is also Banach (because Ker $T$ is closed ) and moreover you have that $$\forall y\in D\,\exists x\in X: Tx=y\Leftrightarrow \overline T([x])=y$$ and $$\|[x]\|=\inf\limits_{z\in Ker T}{\|x-z\|}\leq \|x-0\|=\|x\|\leq C\|y\|$$
Still I can not  finish the proof (with the same constant $C$), because we only get $\forall y\in Y\,\exists x\in X: \|[x]\|\leq C\|y\|$ and $Tx=y$ (to prove this, apply the arguments above but for $\overline T$). But from this I can not conclude that $\|x\|\leq C\|y\|$. However, if we assume that $X$ is reflexive, then $\|[x]\|=\inf\limits_{z\in Ker T}{\|x-z\|}$ attains its infimum at some point $z_0\in Ker T$ because the norm is convex, weakly lower semicontinuous and $Ker T$ is weakly closed. Then we get $\forall y\in Y,\,\exists x\in X, z_0\in Ker T:\,T(x-z_0)=Tx=y$ and $\|x-z_0\|=\|[x]\|\leq C\|y\|$ so the property that you want is proved (with the same constant $C$ for $D$ and $Y$).
EDIT Because it is not said that the constant should be the same for the whole space $Y$ (see my comment under the question), then from the proved above for the operator $\overline T$ (without assuming reflexivity of $X$) we have that  $\forall y\in Y\,\exists x\in X:\,\inf\limits_{z\in Ker T}{\|x-z\|}=\|[x]\|\leq C\|y\|$ and $\overline T([x])=y\Leftrightarrow Tx=y$. Now just increase the constant $C$ by $1$ to get that for each $y$ the quantity $(C+1)\|y\|$ is strictly greater than $\|[x]\|=\inf\limits_{z\in Ker T}{\|x-z\|}\Rightarrow \exists z_0\in Ker T:\, \inf\limits_{z\in Ker T}{\|x-z\|}\leq\|x-z_0\|\leq (C+1)\|y\|$. So the solution for $y$ becomes $x-z_0$ because $T(x-z_0)=y$ and $\|x-z_0\|\leq (C+1)\|y\|$
Note that we also proved that $T$ is surjective, and so an open mapping by the Open mapping theorem.
Finally, we conclude that: 
1)If the space $X$ is in addition reflexive, the constant $C$ stays the same for all the space $Y$.
2)If we do not assume that $X$ is reflexive, the proof for a general (not necesarily injective) operator $T$ is done by considering the operator $\overline T: X/Ker T\to Y$
