Your reasoning seems good, but you have not written the description of $T$. Therefore we can't say if your answer is right or wrong.
But let's go through what you have written. You have talked about a basis of $V $, namely $u_1,u_2,u_3,v_1,v_2$, out of which $u_i$ are in the null space. This directly translates to $T(u_i)=0$.
Now, all you need to do is ensure that $T(v_1)$ and $T(v_2)$ are linearly independent. The best way to do this is to simple consider two linearly independent vectors in $W$, say $w_1$ and $w_2$ (if $\dim W < 2$, you can't construct such a $T$!), and map $v_1 \to w_1$ and $v_2 \to w_2$.
Therefore, your full definition of the operator $T$ is like this: Let $x \in V$ be of the form $x = x_1u_1 + x_2u_2 + x_3u_3 + x_4v_1+x_5v_2$. Then, define $T: V \to W$ by $T(x) = x_4w_1 + x_5w_2$.
I leave you to check that $\dim [\text{null }T] = 3$ and $\dim [\text{range }T] = 2$.
Just a friendly reminder : If $\dim W < 2$, then no such $T$ exists!
To give an explicit example : define $T : \mathbb R^5 \to \mathbb R^5$ by $T(a,b,c,d,e) = (a,b,0,0,0)$. This has rank two, nullity three.
Your supposed definition of infinite dimensional vector space doesn't make sense. A vector space $V$ is infinite dimensional if no finite subset of $V$ generates $V$. This is equivalent to the assertion that $V$ has an infinite subset which is linearly independent.
The space $\mathcal{C}\bigl([a,b]\bigr)$ is infinite dimensional because the set $\{1,x,x^2,x^3,\ldots\}$ is linearly independent.
Best Answer
Here's a slicker way to approach this than the hint suggests, though it's much the same logic underneath.
Suppose $U$ is a subspace of $V$, and consider $S = T|_U : U \to W$. Note that $\operatorname{N}(S) \subseteq \operatorname{N}(T)$ and $S(U) = T(U) \subseteq T(V)$ (if you doubt this at all, you should prove it!).
Now, in particular, if $V$ is an infinite-dimensional vector space, then $V$ contains a finite-dimensional subspace $U$ of any dimension $n$. In particular consider $$n = \operatorname{dim} \operatorname{N}(T) + \operatorname{dim} T(V) + 1,$$ assuming the nullspace and image are finite-dimensional. Then, defining $S$ as the restriction to $U$ as above, the rank-nullity theorem says \begin{align*} \operatorname{dim} U &= \operatorname{dim} \operatorname{N}(S) + \operatorname{dim} S(U) \\ &\le \operatorname{dim} \operatorname{N}(T) + \operatorname{dim} T(V) \\ &= n - 1 = \operatorname{dim} U - 1, \end{align*} a contradiction.