Express each hyperplane by its analytical equation $a_{i1}x_1 + a_{i2}x_2 + \cdots + a_{in}x_n = b_i, i = 1, \ldots, k$. Then the intersection is the solution to the linear system $Ax = b$, i.e., the set $S = \{x \in \mathbb{R}^n: Ax = b\}$, where
\begin{align*}
A = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{k1} & a_{k2} & \cdots & a_{kn}
\end{pmatrix}, \quad
x = \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix}, \quad
b = \begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_n \end{pmatrix}.
\end{align*}
Now you can unleash the theory of linear system to address the problem. When $b \neq 0$, then $S \neq \emptyset$ if and only if $\text{rank}((A, b)) = \text{rank}(A)$. However, in general $S$ is not a subspace of $\mathbb{R}^n$, hence it's pointless to discuss the dimension of $S$.
When $b = 0$, then it is well known that $\dim(S) = n - \text{rank}(A)$. So to determine the dimension of $S$ boils down to determine the rank of the coefficient matrix $A$. Since $0 \leq \text{rank}(A) \leq \min(n, k) = n$, $0 \leq \dim(A) \leq n$.
It seems that (I) is used implicitly when using the words "selecting the maximal subset" there.
[[ We might have to see the whole argument to be certain , though I think this is where the author used (1) ]]
Certain Properties will not allow selecting "selecting the maximal subset" , like this Eg 1 :
Let Set $X=\{1,2,-3,4,-5,6,-7,-8\}$
We want Sub-Set $Y$ where the Sum is Positive.
We can check that $Y$ can be $\{1,2\}$ , $\{1,2,4\}$ , $\{1,-3,4\}$ , $\{-3,4,-5,6\}$ , where Sum is indeed Positive.
Let $Y=\{1,-3,4\}$ , with Positive Sum.
Yet , there are Sub-Sets of $Y$ whose Sum goes Negative too : $\{1,-3\}$ , $\{-3\}$
Hence , we can not talk about "selecting the maximal subset" here.
When we include new element $-5$ to $Y$ , then Sum goes Negative.
Yet , we can add two new elements $-5$ & $6$ to $Y$ , where Sum will remain Positive.
It reiterates that we can not talk about "selecting the maximal subset" here.
Consider this Eg 2 :
We have list of numbers $Z = [+5,-5,+5,-5,+5]$ where we want to pick out "contiguous" numbers with Positive Sum.
We have various ways to try that :
- Only first $+5$
- Only first three numbers $+5,-5,+5$
- Only last $+5$
- Only last three $+5,-5,+5$
- $\cdots$
There is not much meaning to unique maximal subset here too.
[[ I am just making explanatory examples to show that maximal subset may or may not exist , the answer will stand without these examples too ]]
When the author talks about "selecting the maximal subset" with the Property "strongly linearly independent" , then (I) implicitly allows it.
When we have made the selection , it can not occur that we can add new element which makes it lose that Property yet we can add two new elements to retain that Property.
In other words , "selecting the maximal subset" will make sense here.
Best Answer
The book's proof can be expanded to be more explicit: it is, for every subspace $L$, defining a subspace usually known as $L^{\perp}$ perpendicular to the original in the sense that $z\in L^{\perp}$ if and only if for every $x\in L$ we have that the inner product $z\cdot x = 0$ - i.e. that $z$ is "perpendicular" to every $x\in L$.
The idea is that if you let $z_1,\ldots,z_k$ be a basis for $L^{\perp}$, then you can define hyperplanes $H_i$ to be the set of $x$ such that $z_i\cdot x = 0$. The claim is that the fact that $z_1,\ldots,z_k$ is a basis for $L^{\perp}$ implies that $H_1\cap H_2\cap \ldots \cap H_k$ must equal $L$ and therefore that $L$ is a finite intersection of hyperplanes.
For instance, if you had the subspace $L=\{(0,t,t) : t\in\mathbb R\}$ in $\mathbb R^3$, one would find that $L^{\perp}$ could be described as $\{(x,y,-y):x,y\in\mathbb R\}$ which has a basis $\{(1,0,0),(0,1,-1)\}$ which then says that $L$ is the intersection of the two hyperplanes defined by the tuples $(x,y,z)$ such that $x=0$ and $y-z=0$.
This does subtly rely on the nontrivial fact that $L^{\perp\perp} = L$.
Though it's not necessary to understand this proof, the idea behind it is encapsulated in the notion of duality which is basically a method in linear algebra in which one transforms lines (which can be represented by vectors!) into hyperplanes and spanning sets into intersections of hyperplanes. The motivation for the proof in the book is to apply this reasoning (i.e. to look at $L^{\perp}$) to try to get a new statement out of our preexisting knowledge of bases.