Let $T:D(T)\subseteq H_0\to H_1$ be a closed linear operator between the Hilbert spaces $H_0$ and $H_1$. Then we can decompose the Hilbert spaces as $$H_0=\text{ker}(T)\oplus\overline{\text{ran}}(T^*),\, H_1=\overline{\text{ran}}(T)\oplus\text{ker}(T^*)$$ respectively.
With these decompositions in mind, suppose that both $H_0$ and $H_1$ coincide, that is, suppose $H_1=H_0=:H$, and regard in particular the decomposition $$\color{red}{H=\overline{\text{ran}}(T)\oplus\text{ker}(T^*)}$$
Suppose also that $\overline{\text{ran}}(T)\subseteq\text{ker}(C)$, where $C:D(C)\subseteq H\to H$ is another closed linear operator. It then follows, computing that adjoint of $C$, that $H$ can also be decomposed as
$$\color{red}{H=\text{ker}(C)\oplus\overline{\text{ran}}(C^*)}.$$
My question: now compare $\color{red}{\text{these two}}$ decompositions of $H$ simultaneously. Since $\overline{\text{ran}}(T)\subseteq\text{ker}(C)$, is it obvious $\overline{\text{ran}}(C^*)\subseteq\text{ker}(T)$ follows? 'Obvious' here meaning that by visual inspection alone; is it enough to plug the outstanding inclusion chain when comparing the two decompositions? Or, is there some calculation or construction one would need to do in order to properly establish the last inclusion?
My loose thinking is since $\overline{\text{ran}}(T)$ is contained in a potentially larger set, $\text{ker}(C)$, that in order to maintain both decompositions simultaneously, $\overline{\text{ran}}(C^*)$ needs to be contained in the potentially smaller $\text{ker}(T)$, thus yielding the last inclusion. So that, the direction the remaining inclusion should point in be obvious.
In order to be able to define $T^*$ as a single-valued operator we need to assume $T$ is densely defined so in this answer I add that assumption.
First notice that $\ker T^*$ is closed since $T^*$ is a closed operator. This means that it suffices to check that $\operatorname{Ran}T \subseteq \ker C$ implies that $\operatorname{Ran C^*} \subseteq \ker T^*$.
For this, fix $x \in D(C^*)$. Then for $y \in D(T)$, we have that $\langle T^* C^* x, y \rangle = \langle x, CTy \rangle = \langle x, 0 \rangle = 0$. Since $D(T)$ is dense, this implies that $T^*C^*x = 0$ which gives the desired result.