Consider the Hilbert space $X=\mathbb{R}\times L^2[-d,0]$ with norm $|\cdot|$ and consider the bounded linear operator $A \colon X \to X$ defined by $$Ax=\left [x_0+\int_{-d}^0x_1(\xi)d \xi , -\int_{-r}^\cdot x_1(\xi)d \xi \right ]$$ for every $x=[x_0,x_1] \in X$.
Now consider its adjoint operator $A^*$ which is $$A^*x=\left [ x_0,x_0-\int_{\cdot}^{0} x_1(\xi) d \xi \right]$$ for every $x=[x_0,x_1] \in X$.
Finally consider $T \colon X \to X$ defined by $Tx=[x_0,0]=T^*x$ for every $x=[x_0,x_1] \in X$.
Now as we have the following inclusion for the ranges $R(T^*) \subset R(A^*)$ by proposition B.1 in [Da Prato, Giuseppe, and Jerzy Zabczyk. Stochastic equations in infinite dimensions. Cambridge university press, 2014] we have for some $C>0$ (independent of $x$) $$|x_0|=|Tx| \leq C |Ax| $$
But this is in contradiction with the following: consider $|Ax|$, i.e. $$ |A x|=\left|x_{0}+\int_{-r}^{0} x_{1}(s) d s\right|^{2}+\int_{-r}^{0}\left|-\int_{-r}^{\xi} \eta_{1}(s)\right|^{2} d s $$ The latter norm does not control $\left|x_{0}\right|.$ Indeed, consider in $X$ the sequence $$ x^{n}=\left(x_{0}^{n}, x_{1}^{n}\right), \quad x_{0}^{n}=1, x_{1}^{n}=-n \mathbf{1}_{[-1 / n, 0]}(\cdot), \quad n \geq 1 $$ Supposing without loss of generality that $1 / n<r$, we have $$ |A x^{n}|=0+\int_{-\frac{1}{n}}^{0}\left|\int_{-\frac{1}{n}}^{\xi} n d s\right|^{2} d \xi=\int_{-\frac{1}{n}}^{0} n^{2}\left(\xi+\frac{1}{n}\right)^{2} d \xi=\frac{1}{3 n} \longrightarrow 0 . $$ Therefore, we have $|x_{0}^{n}|=1$ and $|A x^{n}| \rightarrow 0$.
Where is the problem? Thanks in advance (Da prato in his book says "we recall some results on linear operator and their images ass they are very important in control theory" letting the reader understand that that is a well-known result).
Your range inclusion $R(T^*) \subset R(A^*)$ fails. Indeed, $(1,0)$ is in the range of $T^*$, but every $(x_0,x_1)$ in the range of $A^*$ satisfies $x_1(0) = x_0$ (note that $x_1$ is continuous).