Here $X$ is Banach space and $T^{\ast}$ is the continuous dual(adjoint) of $T,R(T^{\ast})$ stands for the range of $T^{\ast}.$
The following is what I get:
If such a $T$ exists, then $X$ cannot be reflexive: If so, since $T^{\ast}$ is not surjective, we can find some $0\ne f_0\in X^{\ast}.$ Since $R(T^{\ast})$ is closed, by Hahn-Banach there is some $\eta\in X^{\ast\ast}$ so that $$\left\langle\eta,f_0\right\rangle>0 ~\text{and}~\left\langle\eta,f\right\rangle=0,\forall f\in R(T^{\ast})$$ while $X$ is reflexive, it means exactly that $\eta=Jx_0$ for some $x_0\in X,$ where $J$ is the canonical embedding. Then $$0=\left\langle\eta,f\right\rangle=\left\langle Jx_0,f\right\rangle=\left\langle f,x_0\right\rangle,\forall f\in R(T^{\ast})$$ that implies $\left\langle g,Tx_0\right\rangle=\left\langle T^{\ast}g,x_0\right\rangle=0,\forall g\in X^{\ast}\Rightarrow Tx_0=0\Rightarrow x_0=0$ since $T$ is injective. So $\eta=0,$ a contradiction!
But I cannot construct such a $T$ when $X$ is not reflexive, can someone help me?