There exists a subspace $U$ of $V$ such that $U\cap null\ T=\{0\}$ and $range\ T = \mathcal{J}=\{Tu|u\in U\}$ Is the following Proof Correct?
Theorem. Given that $V$ is finite-dimensional and $T\in\mathcal{L}(V,W)$, there exists a subspace $U$ of $V$ such that $U\cap null\ T=\{0\}$ and $range\  T = \mathcal{J}=\{Tu|u\in U\}$
Proof. Since $V$ is finite-dimensional it follows that all subspaces of $V$ are finite-dimensional then in particular for $null\ T$ there exist a list of vectors $w_1,w_2,...,w_m$ such that they act as a basis for $null\ T$.
Moreover since any linearly-independent list in $V$ can be extended to a basis of $V$. We may choose the vectors $u_1,u_2,...,u_n$ such that following list is a basis for $V$.
$$w_1,w_2,...,w_m,u_1,u_2,...,u_n\tag{1}$$
where $u_1,u_2,...,u_n$ is a basis for $U$.
let $w$ be an arbitrary element in $range\ T$ consequently for some $v\in V$, $Tv=w$ moreover using $(1)$ for some scalars $a_1,a_2,...,a_m,b_1,b_2,...,b_n$
$$w=Tv=T\left(\sum_{j=1}^{m}a_jw_j+\sum_{i=1}^{n}b_iu_i\right) = \sum_{j=1}^{m}a_jTw_j+\sum_{i=1}^{n}b_iTu_i\tag{2}$$
but $\forall j\in\{1,2,...,m\}(Tw_j=0)$ thus we may state $(2)$ as follows
$$w=\sum_{i=1}^{n}b_iTu_i = T(\sum_{i=1}^{n}b_iu_i)\tag{3}$$
thus $w\in\mathcal{J}$ implying that $range\  T\subseteq\mathcal{J}$, it is evident that $\mathcal{J}\subseteq\ range\ T$ thus $range\ T = \mathcal{J}$.
Now consider an arbitrary $u\in U\cap\ null\ T$ it then follows that 
$$u = \sum_{j=1}^{m}a_jw_j=\sum_{i=1}^{n}b_iu_i\tag{4}$$ thus
$$0 = \sum_{j=1}^{m}a_jw_j-\sum_{i=1}^{n}b_iu_i\tag{5}$$
but the list in $(1)$ is a basis for $V$ thus $a_1=a_2=...=a_m=b_1=b_2=...=b_m=0$ implying that $u=0$
$\blacksquare$
 A: There is no need to assume that $V$ is finite dimensional, and we have a more  general Theorem:$\newcommand{\ran}{\operatorname{range}}\newcommand{null}{\operatorname{null}}\newcommand{\e}{\varepsilon}$

The original image of a basis for 
  $\ran T$ and a basis for 
  $\null T$
  together form a basis for $V$.

Proof. Suppose $\{\varepsilon_i\}_{i\in \Gamma_1}$ is a (hamel) basis for $\null T$, $\{\beta_j\}_{j\in \Gamma_2}$ is a basis for $\ran T$, and $\alpha_j=T^{-1}(\beta_{j}), j\in\Gamma_2.$ 
$\forall v\in V,$ suppose $Tv=\sum_{j\in \Gamma_2^0}k_j \beta_j$ for some finite subset $\Gamma_2^0$ of $\Gamma_2$, then $v-\sum_{i\in \Gamma_2^0}k_j\alpha_j\in \null T$, so there exists some finite subset $\Gamma_1^0$ of $ \Gamma_1$ such taht $v-\sum_{j\in\Gamma_2^0}k_j \alpha_j=\sum_{i\in\Gamma_1^0}k_i\e_i$. Hence every vector in $V$ is a linear combination of $\{\e_i\}_{i\in \Gamma_1}\cup\{\alpha_j\}_{j\in\Gamma_2}$.
Suppose $\sum_{i\in\Gamma_1^0}k_i\e_i+\sum_{j\in\Gamma_2^0}k_j \alpha_j=0$ for some finite subset $\Gamma_1^0$ and $\Gamma_2^0$ of $\Gamma_1$ and $\Gamma_2$,respectively. Let $T$ act on both side, we have $\sum_{j\in\Gamma_2^0}k_j \beta_j=0$, thus $k_j=0(j\in \Gamma_2)$. Hence $\sum_{i\in\Gamma_1^0}k_i\e_i=0$ and $k_i=0(i\in\Gamma_1)$. Therefore, $\{\e_i\}_{i\in \Gamma_1}\cup\{\alpha_j\}_{j\in\Gamma_2}$ is linear independent.$\blacksquare$
Back to your question, $U=\operatorname{span}\{\alpha_j\}_{j\in \Gamma_2}$ is required.
