Prove $\operatorname{Rank} (T) +\operatorname{Nullity} (T) = \dim V$ 
Suppose that $T : V \to W$ is a linear transformation from a finite dimensional vector space $V$ to a vector space $W$. Then $\operatorname{rank} (T)$ is finite and $\operatorname{Rank} (T) +\operatorname{Nullity} (T) = \dim V$.

Here's my proof so far:

Let $n = \dim V$ and $r = \operatorname{rank}(T)$, so we must prove $r = n - \dim \ker(T)$. suppose first that $1 \le r \le n-1$. Let $S = (v_1,\ldots,v_m)$ be a basis for $\ker T$, and choose vectors $T = (v_{m+1},\ldots,v_n)$ in $V$ so that $S \cup T = (v_1,\ldots,v_n)$ is a basis for $V$. we now claim that $U = (Tv_{m+1},\ldots,Tv_n)$ is a basis for range $T$, and with this done we will have $r = \operatorname{rank} T = \dim \operatorname{range} T = \#U = n-m = n - \dim \ker T$.

And from here, I have no idea what to do. My classmate told me that I have to show $U$ is linearly independent and that $\operatorname{Span} U = \operatorname{range} T$ (?) but I'm not quite sure$\ldots$
Am I approaching this theorem too complicatedly? Any help would be very helpful!!

Well, you are indeed almost done: a basis is a linear independent generating
set, so as your classmate mentioned, you only need to check these two
properties for $v_{m+1},\ldots,v_n$.
For linear independence assume
$$0 = \sum_{i=m+1}^n T(v_i)\lambda_i = T(\sum_{i=m+1}^n v_i\lambda_i)$$
and see what this would mean for the $v_{m+1},\ldots,v_n$.
For generation, given a typical element $T(v)$ of $\operatorname{range}(T)$,
express $v$ with the basis of $V$ and observe what happens to
the $v_1,\ldots,v_m$ when $T$ is applied.
