- It is true that $\textbf{u}$ and $\textbf{v}$ are linearly independent with these assumptions, however, it is not sufficient to claim it based on intuition. You can show it as follows:
Suppose $\alpha\textbf{u} + \beta\textbf{v} = \textbf{0}$. Then
$$
0 = \langle \textbf{v}, \textbf{0}\rangle = \langle \textbf{v}, \alpha\textbf{u} + \beta\textbf{v} \rangle = \alpha \langle \textbf{v}, \textbf{u}\rangle + \beta\langle\textbf{v}, \textbf{v}\rangle = 0 + \beta|\textbf{v}|^2.
$$
You can conclude from here that $\beta = 0$ (why?). A similar calculation shows that $\alpha = 0$, from which you can conclude that $\textbf{u}$ and $\textbf{v}$ are linearly independent
- For part two it should be easy to come up with a counter example in $\mathbb{R}^2$ to find two linearly independent vectors that are not orthogonal. This will show the statement is false.
To expand on Cameron Williams' comment:
The parallelogram law is given by
$$2||x||^2+2||y||^2=||x+y||^2+||x-y||^2\text{ f.a. }x,y\in V$$
for a vector space and it can be shown that every norm induced by an inner product satisfies this property.
However, taking your norm, we find that for $x=(1,0),y=(0,-1)$, we have
$$2||x||^2+2||y||^2=2(|1|+|0|)^2+2(|0|+|-1|)^2=2(1)^2+2(1)^2=2+2=4$$
but
$$||x+y||^2+||x-y||^2=(|1+0|+|0-1|)^2+(|1+0|+|0+1|)^2=(|1|+|-1|)^2+(|1|+|1|)^2=2^2+2^2=4+4=8$$
EDIT: I want to give you a proof that every inner-product induced norm satisfies the parallelogram law:
Let $||x||:=\sqrt{\langle x,x\rangle}$ for an inner product $\langle\cdot,\cdot\rangle$ on a real vector space $V$. Thus $||x||^2=\langle x,x\rangle$ and followingly:
$$||x+y||^2=\langle x+y,x+y\rangle=\langle x,x\rangle +\langle x,y\rangle+\langle y,x\rangle+\langle y,y\rangle$$
and
$$||x-y||^2=\langle x-y,x-y\rangle=\langle x,x\rangle -\langle x,y\rangle-\langle y,x\rangle+\langle y,y\rangle$$
Thus
$$||x+y||^2+||x-y||^2=\langle x,x\rangle +\langle x,y\rangle+\langle y,x\rangle+\langle y,y\rangle+\langle x,x\rangle -\langle x,y\rangle-\langle y,x\rangle+\langle y,y\rangle$$
i.e.
$$||x+y||^2+||x-y||^2=2\langle x,x\rangle+2\langle y,y\rangle=2||x||^2+2||y||^2$$
The so called polarization identity establishes the converse, i.e. it shows that for a normed space $(V,||\cdot||)$, if $||\cdot||$ satisfies the parallelogram law, then it comes from an inner product, i.e. there is an inner product $\langle\cdot,\cdot\rangle$ on $V$ s.t. $||x||^2=\langle x,x\rangle$ f.a. $x\in V$.
Best Answer
I too found the "proof" given in the text of the question obscure and confusing, so much so that I can't really say more; hence my own work is presented below; it is not difficult.
To avoid technical difficulties I assume $T$ is a bounded operator on $V$.
These things being said, let
$x \in N(T); \tag 1$
then
$Tx = 0, \tag 2$
whence
$\forall y \in V, \: \langle y, Tx \rangle = 0; \tag 3$
thus
$\forall y \in V, \; \langle T^\dagger y, x \rangle = 0, \tag 4$
that is,
$x \in R(T^\dagger)^\bot, \tag 5$
and thus
$N(T) \subset R(T^\dagger)^\bot; \tag 6$
also, in the event that (5) binds, we have
$\forall y \in V, \; \langle T^\dagger y, x \rangle = 0, \tag 7$
or
$\forall y \in V, \; \langle y, Tx \rangle = 0, \tag 8$
which immediately yields
$Tx = 0 \Longrightarrow x \in N(T), \tag 9$
and hence
$R(T^\dagger)^\bot \subset N(T); \tag{10}$
finally, (6) and (10) together imply
$R(T^\dagger)^\bot = N(T), \tag{11}$
$OE\Delta$.