Source:
I think the important parts (as mentioned in the question) about Section 2 is about dot products and projections, specifically $x = x^{||}+x^\perp$.
Proof attempt:
$\underline{\text{Define }w_2}$
First, let $w_1=v_1$. Since $v_2 = v_2^{\parallel}+v_2^{\perp}$ (projection of $v_2$ on $v_1$ + projection of $v_2$ on a vector orthogonal to $v_1$) , we can declare $w_2 =v_2^{\perp}=v_2 - v_2^{\parallel}$ as $w_2$ is by definition orthogonal to $v_1$.
$\underline{\text{Show that Span}(w_1, w_2) = \text{ Span}(v_1,v_2)}$
$\underline{\text{Show that Span}(w_1, w_2) \subseteq \text{ Span}(v_1,v_2)}$
Let $w'\in\text{Span}(w_1, w_2)$. Then, $$\begin{align*}w'&=c_1 w_1+c_2 w_2\\&=c_1 v_1+c_2 (v_2-v_2^{\parallel})\\&=(c_1-\frac{v_2\cdot v_1}{||v_1||^2})\cdot v_1+c_2 (v_2)\in\text{ Span}(v_1, v_2)\end{align*}$$
$\underline{\text{Show that Span}(w_1, w_2) \supseteq \text{ Span}(v_1,v_2)}$
Let $v'\in\text{Span}(v_1, v_2)$. Then, $$\begin{align*}v'&=c_1 v_1+c_2 v_2\\&=c_1 v_1+c_2 (v_2^{\perp}+v_2^{\parallel})\\&=(c_1+\frac{v_2\cdot v_1}{||v_1||^2})\cdot v_1+c_2 (w_2)\in\text{ Span}(w_1, w_2)\end{align*}$$
So $\text{Span}(w_1, w_2) = \text{ Span}(v_1,v_2)$.
$\underline{\text{Where I'm stuck}}$
Let $w_n = v_n - \sum_{i=1}^{n-1}\frac{v_n\cdot v_i}{||v_i||^2}$. I'm not sure how to proceed from here as it seems like I'd have to prove it for $k-1$ times...