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 $\overline T: X/Ker T\to Y$