I start with a definition $\vec u \cdot \vec v = \Vert u\Vert \Vert v \Vert \cos\theta$
Then, using this I say $proj_\vec v(\vec u) := (\vec u \cdot \hat v)\hat v = \vec u_\vec v$
$$ \implies proj_\vec v(\vec u) \cdot \vec v = (\vec u \cdot \hat v)\hat v \cdot \vec v = (\vec u \cdot \hat v)\Vert v \Vert \cos 0 = \vec u \cdot \vec v $$$$ \therefore \vec u \cdot \vec v = \vec u_\vec v \cdot \vec v \quad\forall \vec u,\vec v \quad\epsilon\space V \implies \vec u - \vec u_\vec v=\vec u_{\vec v_{\bot}},\qquad \vec u_{\vec v_{\bot}}\cdot\vec v = 0 $$$$ Let \quad\vec u = \vec u_0 + \vec u_1 = \vec u_\vec v + \vec u_{\vec v_{\bot}} = (\vec u_{{\vec v}0} + \vec u_{{\vec v}1}) + (\vec u_{\vec v_{\bot}0} + \vec u_{\vec v_{\bot}1}) $$ The latter components aren't necessarily indicative of whether they are components of V0 or V1 (although pretty obvious), which is what I'm trying to prove without just saying so.
Define $\vec u_0 = \vec u_{{\vec v}0} +\vec u_{\vec v_{\bot}0}$ and $\vec u_1=\vec u_{{\vec v}1}+ \vec u_{\vec v_{\bot}1}$ such that $proj_\vec v(\vec u_{i=\{0,1\}})=\vec u_{{\vec v}i}$
It then follows that as both components (proj and orthogonal) are independent from one another any combination of the orthogonal components will itself be orthogonal to v, $$\vec u_\vec v \not=a\vec u_{\vec v_{\bot}0}+b\vec u_{\vec v_{\bot}1}\quad\forall\ a,b \ \epsilon\ \mathbb{R} \qquad \implies \quad \vec u_\vec v = \vec u_{{\vec v}0}+\vec u_{{\vec v}1}$$ $$ \therefore \vec u \cdot \vec v = \vec u_\vec v \cdot \vec v = (\vec u_{{\vec v}0}+\vec u_{{\vec v}1}) \cdot \vec v = ((\vec u_{{\vec v}0} \cdot \hat v)\hat v+(\vec u_{{\vec v}1} \cdot \hat v)\hat v)\cdot\vec v $$$$ = ((\vec u_0 \cdot \hat v)+(\vec u_1 \cdot \hat v))\hat v \cdot\hat v\Vert v\Vert=(\vec u_0 \cdot \hat v)\Vert v\Vert+(\vec u_1 \cdot \hat v)\Vert v\Vert=\vec u_0 \cdot \vec v+\vec u_1 \cdot \vec v $$ $\quad$ $$ \therefore\qquad \forall\quad \vec u = \vec u_0 + \vec u_1 \ and\ \vec v \ \epsilon \ \space V,\vec u \cdot \vec v = \Vert u\Vert \Vert v \Vert \cos\theta = \vec u_0 \cdot \vec v+\vec u_1 \cdot \vec v $$ (consequently)
$$ let\ \vec u := a\hat i+b\hat j+c\hat k,\ \vec v:=\alpha\hat i+\beta\hat j+\gamma\hat k $$ \begin{align} Then, \vec u\cdot\vec v & = a\hat i\cdot \alpha\hat i+a\hat i\cdot \beta\hat j+a\hat i\cdot \gamma\hat k+b\hat j\cdot \alpha\hat i+b\hat j\cdot \beta\hat j+b\hat j\cdot \gamma\hat k+c\hat k\cdot \alpha\hat i+c\hat k\cdot \beta\hat j+c\hat k\cdot \gamma\hat k \\ \vec u\cdot\vec v & = a\alpha(\hat i\cdot \hat i)+0+0+0+b\beta(\hat j\cdot \hat j)+0+0+0+c\gamma(\hat k\cdot \hat k) \\ & = a\alpha+b\beta+c\gamma \end{align}