Proof of the Riesz Representation Theorem Theorem: Let $F$ be a continuous linear functional on the Hilbert space $H$, then $\exists !$ (exists one and only one) $y \in H$ such that $F(x) = (x,y)$ for $x\in H$.
Proof: 


*

*Uniqueness: $$F(x)=(x,y)=(x,y') \Rightarrow (x,y-y')=0 \space \forall x \in H \Rightarrow ||y-y'||^2=0$$

*Existence: if for any $x\in H, \space F(x) =(x,y)= 0 $ we'll set $y=0$. Let that $F \neq 0. $ and define a closed subspace $H_0 = \{x \in H: F(x)=0\}$. We know that $H \neq H_0$ because $H_0 = F^{-1}(\{0\})$.
Q1: "$H_0$ is a subspace by the linearity of $F$, closed by continuity of $F$." What does this mean? I understand what's a subspace, but don't understand what does it have to do with linearity and continuity.
$\exists z\in H$ such that $||z||=1$. For $u \in H_0, \space F(u)=(u,z) =0$. 
Q2: Does this imply that $u \perp z \space \forall u \in H_0, \forall z \in H$?
Let's define $u=F(x)z - F(z)x, \space x \in H$ 
$$F(u)=(F(x)z - F(z)x,z)=F(x)F(z)-F(z)F(x)=0=F(x)(z,z)-F(z)(x,z)$$ 
Since we know that $||z|| = \sqrt{(z,z)}=1$, then 
$$0=F(x)(z,z)-F(z)(x,z) \Rightarrow F(x) =F(z)(x,z) = (x,\bar{F(z)z)} $$ 
By defining $y= \bar{F(z)}z$ we've completed the proof. 
By the way, the bar on the top denotes the complex conjugate, as is common in Hilbert spaces. i.e. $\alpha (x,y)= (\alpha x,y)=(x,\bar{\alpha} y) $
Corollary: If $H=L^2(\mu)$ and $F$ is a continuous linear functional on $L^2(\mu)$, then $\exists ! \space g \in L^2(\mu)$ such that $$F(f)=\int_X f \bar{g} d\mu = (f,g)$$
In the context of the Radon-Nikodym theorem, $g$ is the R-N derivative.
 A: Question 1: You defined $H_0 = F^{-1}(\{0\})$, where $F \in H^*$ (the continuous linear functionals on $H$). In order for $H_0$ to be a subspace, you have to check that $\lambda x + \mu y \in H_0$ for all $x,y \in H_0$, $\lambda, \mu \in \mathbb K$, right? But 
$$ \lambda x + \mu y \in H_0 \iff F(\lambda x + \mu y) = 0 $$
As $F$ is linear (and $x,y \in H_0$), we have
$$ F(\lambda x + \mu y) = \lambda F(x) + \mu F(y) = 0 $$
So $H_0$ being a subspace follows from the linearity of $F$. In order for $H_0$ to be closed, we must show that for all sequences $(x_n)$ in $H_0$ converging to some $x \in H$ we must have $x\in H_0$. But, as $F$ is continuous, we have that $x_n \to x$, implies $F(x_n) \to F(x)$. As $F(x_n) = 0$ for every $n$, we get $F(x) = 0$, so $x \in H_0$. So the closedness of $H_0$ follows from the continuity of $F$.
Question 2: No, it doesn't. As $H_0$ is a proper closed subspace of $H$ ($H_0 \subsetneq H$ as $F \ne 0$), there is some $z \in H$, with $z \mathrel\bot H_0$ and $\|z\| = 1$. This means that for this $z$, not for all we have 
$$ (z, u) = 0, \quad \text{all } u \in H_0. $$
A: 1) As with any other (linear) vector space, the kernel of a linear transformation is always a vector subspace of the definition domain vector space, and in this case we have that $\;H_0:=\ker F\;$ , with $\,F:H\to\Bbb F\;,\;\;\Bbb F=\Bbb R\;\,\;\Bbb C\,$ . This is also denoted by $\,F\in H^*:=$ the dual space of $\,H\,$ 
The conditions about closedness and being a subspace of Hilbert space then follow from conditions other than the linear ones (i.e., continuity of a linear functional), which hopefully are already clear.
(2) About your second question: the answer's no, since the only elements that is orthogonal to all the elements in the space is the zero vector.
What is written there follows from the fact that $\,H_0\neq H\,$ (in fact, since $\,F\,$ is a non-trivial functional we have that $\;H_0\;$ is a maximal proper subspace of $\,H\,$ . This much is true also in any general vector space ,whether dinite or infinite dimensional over its definition field) . and then $\,H_0^\perp\neq0\;$ , and from here that there exists such $\,z\in H\,$ as implied just before your second question
