Your proof has a huge problem !
Proof: Let $(v_1,\ldots ,v_{n})$ be a basis of $V$, and $(w_1,\ldots ,w_{m})$ be a basis of $W$, such that $$ Tv_{j}=w_{j} $$
for $1\le j\le \text{dim range } T$, and for $\text{dim range } T < j \le n$, $$ Tv_{j}=0. $$
Yes these bases would work, but how do you know they exist ? You haven't proved that ! In fact your proof could be reformulated as
Proof: Let $(v_1,\ldots ,v_{n})$ be a basis of $V$, and $(w_1,\ldots ,w_{m})$ be a basis of $W$, such that the entries of the matrix of $T$ in row $j$, column $j$ are 1, and the rest are 0, for $1\le j\le \text{dim range } T$.
The huge problem is that you're asked to prove that something exists and the first step in your proof is "Let this thing (whose existence I'm trying to prove) exist."
In even simpler terms your proof says
Proof : Assume the theorem is true. Then the theorem is true QED.
Which isn't a proof at all !!!
Let $ v_1,...,v_k,v_{k+1},...,v_{k+s} = v_m$ be a basis of $V$ where $v_{k+1},...,v_{k+s}$ is a basis of $\ker T$. Notice that this exists because we can take a basis of $\ker T$ and extend it to a basis of $V$.
Let $v \in V$. Then there are constants $\lambda_i$ such that
$$v = \sum_{i = 1}^k \lambda_iv_i + \underbrace{\sum_{j = k+1}^{n} \lambda_jv_j}_{\in \ker T} $$
Applying $T$ on both sides and using linearity we have
$$ T(v) = T(\sum_{i = 1}^k \lambda_iv_i) + 0 = \sum_{i = 1}^k \lambda_i T(v_i).$$
Therefore $T(v_1),...,T(v_k)$ spans the range of $T$. We now that show that it is also linearly independent.
We have
$$ \sum_{i=1}^k \lambda_i T(v_i) = 0 \iff T\left( \sum_{i=1}^k \lambda_i v_i\right) = 0 \iff \sum_{i = 1}^k\lambda_iv_i \in \ker T$$
so $\lambda_i = 0$ for all $i$.
Let $w_1 = T(v_1),...,w_k=T(v_k)$ basis of range $T$ and extend it to a basis $w_1,...,w_m$ of $W$. Then $v_1,...,v_n$ and $w_1,...,w_m$ are bases of $V$ and $W$ respectively and the statement is proved.
