Here $T:V \rightarrow W$ and $S:W \rightarrow V$, where V and W are finite dimensional inner-product spaces over the same field. It's also given that $TS=0$.
The question is to prove that $rank(S^*) \geq nullity(T^*)$.
My attempt: Using the fact that $TS=0$, I derived that $rank(S) \leq nullity(T)$. Does the inequality asked automatically follow from this? If not, please give alternate ways to prove the same.
It is not true that $\operatorname{Rank}(S) \leq \operatorname{Nullity}(T)$ implies that $\operatorname{Rank}(S^*) \geq \operatorname{Nullity}(T^*)$. For a counterexample, let $V = W$ be vector spaces of positive dimension and let $S = T = 0$. Then $\operatorname{Rank}(S) = 0$ and $\operatorname{Nullity}(T) = \operatorname{dim}(V) > 0$, but $S^* = T^* = 0$, so also $\operatorname{Rank}(S^*) = 0 < \operatorname{dim}(V) = \operatorname{Nullity}(T^*)$.
The same counterexample actually shows that it cannot follow from $TS = 0$ that $\operatorname{Rank}(S^*) \geq \operatorname{Nullity}(T^*)$, so it appears that the question you are trying to prove is incorrect.
Something that you can prove though is that $\operatorname{Nullity}(S^*) \geq \operatorname{Rank}(T^*)$. This follows from what you did in your attempt by observing that $S^*T^* = (TS)^* = 0$.