If rank$(T)\le$rank$(T^3)$, then intersection of range and null space of $T$ is zero

well first of all, lets consider $T$ as a linear map (this makes stuff in my opinion often a lot clearer). We know that $rank(T)=\dim \textrm{im}(T)$, But we know that $\text{im}(T^3)=\textrm{im}(T\mid_{\textrm{im}(T^2)})$, in particular $\textrm{im}T^3\subset \textrm{im} T$ but that just means in dimensions that $\dim \textrm{im}(T) \ge \dim \textrm{im}(T^3) $ in particular we get with your inequality $$\dim \textrm{im}(T) = \dim \textrm{im}(T^3)$$ But this just means that $T\mid_{\textrm{im}(T)}$ is an isomorphism. So $T\mid_{\textrm{im}(T)}$ has kernel zero. But this immediately gives $$0=\ker(T\mid_{\textrm{im}(T)}) = \ker(T)\cap \textrm{im}{T}$$

Given that, $\operatorname{Rank}(T)\leq \operatorname{Rank}(T^3)\implies n-\operatorname{Rank}(T)\geq n-\operatorname{Rank}(T^3)$
By, Rank-nullity theorem $$\operatorname N(T)\geq \operatorname N(T^3) \tag{1} \label{eqn1}$$ We know that, $$\operatorname{Null space}\,(T)\subseteq \operatorname{Null space}\,(T^3) \tag{2} \label{eqn2}$$ $$\implies \operatorname N(T)\leq \operatorname N(T^3) \tag{3} \label{eqn3}$$ Now, from \eqref{eqn1} and \eqref{eqn3} we have $$\operatorname N(T)=\operatorname N(T^3) \tag{4} \label{eqn4}$$ By \eqref{eqn2} and \eqref{eqn4} we have, $$\operatorname{Null space}\,(T)=\operatorname{Null space}\,(T^3) \tag{5} \label{eqn5}$$ Now, let $x\in \operatorname{Null space} (T) \cap \operatorname{Range} (T)$ then, $T(x)=0$ and $\exists\,\alpha\in V$ such that, $T(\alpha)=x$ Therefore, $T^2(\alpha)=0\implies T^3(\alpha)=0$.
Now, from \eqref{eqn5} we have $\alpha\in$ Null space $(T)\implies T(\alpha)=0=x$
$\square$