Question regarding the simple proof involving dual space and annihilator 
If $\eta$ is a suspace in a finite-dimensional vector spce $v$, then
$\eta^{00}=\eta$
Proof: Observe that we use here the convention, established at the end
of 16, that identifies $v$ and $v''$. By definition, $\eta^{00}$ is
the set of all vectors $x$ such that $[x,y]=0$ for all $y$ in
$\eta^{0}\dots$

16 explains that in convention, we can identify $v$ with its double dual space $v''$, because they are isomorphic. But I'm not sure how the second sentence in the above proof follows.
From my understanding,
by definition, $\eta^{00}$ is the set of all vectors $x$ in $v''$ such that $[y, x]=0$ for every $y\in\eta^{0}$. I'm not sure how we can switch the position of $x$ and $y$ from the isomorphism of $v$ and $v''$.

I believe it is probably this. The Convention you are talking about is to identify $x\in v$ with a functional $\overline{x}:v'\to k$ such that $\overline{x}(y)=y(x)$ for every $y\in v'$. That means that
$$[y,\overline{x}]=\overline{x}(y)=y(x)=[x,y]$$
Now, $\eta^{00}$ is the set of all vectors $\overline{y}\in v''$ such that $[x,\overline{y}]=0$ for all $x\in\eta^0$. (Note we can always see those vectors as being of the form $\overline{y}$ for some $y\in v$, because the correspondence $v\mapsto v''$ happens to be "onto".)
Swap the role of the letters $x$ and $y$, and you get that $\eta^{00}$ is the set of all vectors $\overline{x}\in v''$ such that $[y,\overline{x}]=0$ for all $y\in\eta^0$.
Now, identify $\overline{x}$ with $x$ and you get that $\eta^{00}$ is the set of all vectors $x\in v$ such that $[x,y]=0$ for all $y\in\eta^0$.
