How can I prove
Let $A$ be $n \times n$ with $n$ even, $\text{rank}(A)=n/2$ and $A^2=0$. Then $C(A) = N(A)$.
I thought about proving it by showing that each set is contained in the other. Indeed the one direction is easy.
Let $\mathbf{b} \in C(A)$ then $\mathbf{b}=A\mathbf{x}$ for some $\mathbf{x} \in \Bbb R^n$. Because $A^2=\mathbf{0}$ we get $A\mathbf{b} = AA\mathbf{x} = A^2\mathbf{x}=\mathbf{0}$ and so $\mathbf{b} \in N(A)$.
But what about the other direction?
Let $\mathbf{b} \in N(A)$ then $A\mathbf{b}=\mathbf{0}$. How do I conclude that there exists $\mathbf{x} \in \Bbb R^n$ such that $A\mathbf{x}=\mathbf{b}$?
It does not look right, for instance take $$ A = \begin{bmatrix} 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ \end{bmatrix} $$ where $\dim(C(A)) = 1$ and $\dim(N(A)) = 3$.
(The zero matrix would have also sufficed, but I thought that would have been too trivial an example.)
PS OP has already proved that if $A^{2} = 0$, then $C(A) \subseteq N(A)$. Now the usual range-nullity theorem tells you that $$ n = \dim(C(A)) + \dim(N(A)). $$ Therefore $C(A) = N(A)$ iff their common dimension is $n/2$ iff $\dim(C(A))$, which is the rank of $A$, is $n/2$.