Prove: $T \in L(V,V), T^2 = 0 \iff T(v) \subset n(T)$
Is the following correct?
Proof: $\rightarrow$
Let $T^2 = 0 \iff T(T(v)) = 0$
Suppose $x \in T(v)$ we must show that $x \in n(T) \iff T(x) = 0$
$x\in T(v)$ implies there exists $v\in V$ s.t. $T(v)=x$
Consider $T(T(v))$:
$T(T(v)) = T(x) = 0$
$\leftarrow$ Let $T(v) \subset n(T)$ Suppose $x\in T(v)$. Then we know $x\in n(T) \iff T(x) = 0$. We also know there exists $v\in V$ so that $T(v) = x$.
Consider $T(T(v))$:
$T(T(v)) = T(x) = 0$
Your proof is correct in spirit but it lacks clarity. You need to take care in distinguishing $T(V)$ which is a set (the image of $T$) and $T(\mathbf{v})$ which is a single vector (the image of $\mathbf{v}$ under $T$).
I will rewrite your proof, following your exact logic. Hopefully you will find this version more clear.
Theorem: Let $T:\ V\rightarrow V$ be a linear mapping. Then $T^2 = 0 \iff T(V) \subseteq \operatorname{null}(T)$.
Proof: ($\Rightarrow$) Suppose that $T^2 = 0$. Then for all $\mathbf{v} \in V$ we have $$T^2(\mathbf{v}) = T(T(\mathbf{v})) = \mathbf{0}$$ This shows that $T(\mathbf{v}) \in \operatorname{null}(T)$ for all $\mathbf{v} \in V$ and therefore $T(V) \subseteq \operatorname{null}(T)$.
($\Leftarrow$) Conversely, suppose that $T(V) \subseteq \operatorname{null}(T)$. This means that for all $\mathbf{v} \in V$ we have $$T(\mathbf{v}) \in \operatorname{null}(T) \implies T(T(\mathbf{v})) = \mathbf{0}$$ Therefore $T^2 = 0$. $\square$