Let $F:V\rightarrow U$ be a linear transformation. We have to show that the preimage of any subspace of $U$ is a subspace of $V$.
My proof:
Say $W$ is a subspace of $Im(F)$ and $T$ is a subspace of $V$. Now consider $u,w\in W$, such that there exist $v,p\in T$, such that $F(v)=u, F(p)=w$. $u+w\in W$ as it is a subspace, $$u+w=F(v)+F(p)=F(v+p)$$ We have this by linearity. Now $v+p\in T$, so the preimage of $W$ is a subspace. $F(0)=0$. $0\in W,T$ as these are subspaces.
Proof verification.
As written, your answer is wrong as it is possible that a subspace of $U$ is not contained $Im(T)$ (i.e. whenever $T$ is not surjective).
You have to start with a subspace of $U$, not one that is a subspace of $Im(T)$.
As promised, here is a proof.
Let $W$ be a subspace of $U$. We will show that $F^{-1}(W)$ is a subspace of $V$.
Since $F(0) = 0 \in W$, we know that $0 \in F^{-1}(W)$.
Now, let $x,y \in F^{-1}(W) := \{v \in V: Fv \in W\}$, let $\lambda \in \mathbb{F}$. Then $Fx, Fy \in W$, and hence $F(x + \lambda y) = F(x) + \lambda F(y) \in W$, as $W$ is a subspace. It follows that $x + \lambda y \in F^{-1}(W)$, and we can conlude that $U$ is indeed a subspace of $V$.