Characterization for invariant subspace

You already have a lot of the crucial ideas laid out! The relation between part 3 and part 2 is seems to be that you are exchanging everything with a dual-ish concept, so to speak: we're looking at images instead kernels, and rather than $S$ being one-dimensional, the orthogonal complement $S^\bot$ is one-dimensional. So, somehow once we know part 2, part 3 should not be too far away.

So, assume the conditions of part 3 and that $S$ is $T$-invariant. By part 1, this means $S^\bot$ is $T^*$-invariant. But since $\dim(S^\bot) = \dim V - \dim S = 1$, we can apply part 2 to $T^*$ and $S^\bot$, and we find that this is equivalent to there being an eigenvalue $\lambda$ of $T^*$ so that $S^\bot \subset \ker (T^* - \lambda I)$. But if you know the relation $$ W \subset U \implies U^\bot \subset W^\bot,$$ then we can deduce $\ker(T^* - \lambda I)^\bot \subset (S^\bot)^\bot = S.$ And you've already indicated that you know that $\text{im}(T - \lambda I) \bot \ker((T - \lambda I)^*)$. Hence $$ \text{im}(T - \lambda I) \subset \ker((T - \lambda I)^*) = \ker(T^* - \lambda I) \subset S.$$

This proves the one direction, and the other direction goes similarly by working backwards if you know that $\ker(T)$ and $\text{im}(T^*)$ are not only orthogonal, but that they are indeed even orthogonal complements of one another, i.e. applying $\bot$ to one gives you the other.

All of these relations I am using are fairly elementary properties of orthogonal complements in finite dimensions, so I'm kinda hoping you already know about them, but if that's not the case, it might be a good exercise to try and prove them yourself! Hope that helps, let me know if something is still unclear!

EDIT: I indeed left out an argument why, in the above, $\lambda$ is an eigenvalue of $T$, we only know it is an eigenvalue of $T^*$. This part can be tricky if you approach it from the wrong angle; an eigenvector of $T^*$ will in general not be an eigenvector of $T$, so there is no way to directly compare things. However, there is still hope for an elementary proof.

We know that $\text{im}(T - \lambda I) \subset S$. But that means $T - \lambda I$ can be understood as a map from the 3-dimensional space $V$ into the 2-dimensional space $S$. Hence, simply by the rank-nullity theorem, it must have a nontrivial kernel, so there is a $v \in V$ with $Tv - \lambda v = 0$. But that must be exactly your eigenvector!