Proof verification

You have proved that the solution set of the system $Ax = 0$ is a subspace of $F^n$. What the question asks you to do is show that any subspace is the solution set of some system of equations. That is, you have to show that, given a subspace $W$ of $F^n$, there is a matrix $A$ such that $\{x \in F^n \mid Ax = 0\} = \operatorname{Nul}A = W$.

Disclaimer: Not entirely sure about this.

Proof: Let $V = \mathbb{F^{n}}$. Let $W$ be a subspace of $V$. By definition, every subspace of $V = \mathbb{F}^n$ must contain the $0$ vector and must be closed under addition and scalar multiplication. Therefore, all linear combinations of the system $Ax = a_1x_1+\cdots+a_nx_n = 0$ are contained in $W$, so it is clear that $AX = 0 \in W$, and as a result, $W$ can be represented as the solution space to the homogenous equation. $\Box$

Let $W$ be your subspace. If $W$ is all of the space, it is the kernel of the null matrix. If it is $\{0\}$, it is the kernel of any nonsingular matrix. So assume your subspace is proper, and let $\{w_1,\ldots,w_s\}$ be a basis. Extend this to a basis $\{w_1,\ldots,w_s,w_{s+1},\ldots,w_n\}$ of your space, let $\{w_{s+1},\ldots,w_n\}=W'$. Define the transformation $f:V\to V$ as the projection onto $W'$ through $W$. Then $W=\ker f$, and you can consider the matrix of $f$ in the canonical basis to get an appropriate system of equations.

Alternatively, use the canonical inner product: if $W$ is your subspace, consider $W^\perp$. Then use that $W=(W^\perp)^{\perp}$ to get a set of equations for $W$, namely of the form $w_i\cdot x=0$ for $i=1,2,\ldots, r$ where $\{w_1,\ldots,w_r\}$ is a base for $W^{\perp}$.